Design Examples Aisc 2016.pdf

COMPANION TO THE AISC STEEL CONSTRUCTION MANUAL Volume 1: Design Examples

Version 15.1

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AISC © 2019 by American Institute of Steel Construction

All rights reserved. This publication or any part thereof must not be reproduced in any form without the written permission of the publisher. The AISC logo is a registered trademark of AISC. The information presented in this publication has been prepared following recognized principles of design and construction. While it is believed to be accurate, this information should not be used or relied upon for any specific application without competent professional examination and verification of its accuracy, suitability and applicability by a licensed engineer or architect. The publication of this information is not a representation or warranty on the part of the American Institute of Steel Construction, its officers, agents, employees or committee members, or of any other person named herein, that this information is suitable for any general or particular use, or of freedom from infringement of any patent or patents. All representations or warranties, express or implied, other than as stated above, are specifically disclaimed. Anyone making use of the information presented in this publication assumes all liability arising from such use. Caution must be exercised when relying upon standards and guidelines developed by other bodies and incorporated by reference herein since such material may be modified or amended from time to time subsequent to the printing of this edition. The American Institute of Steel Construction bears no responsibility for such material other than to refer to it and incorporate it by reference at the time of the initial publication of this edition. Printed in the United States of America

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PREFACE The primary objective of this Companion is to provide guidance and additional resources of the use of the 2016 AISC Specification for Structural Steel Buildings (ANSI/AISC 360-16) and the 15th Edition AISC Steel Construction Manual. The Companion consists of design examples in Parts I, II and III. The design examples provide coverage of all applicable limit states, whether or not a particular limit state controls the design of the member or connection. In addition to the examples that demonstrate the use of the AISC Manual tables, design examples are provided for connection designs beyond the scope of the tables in the AISC Manual. These design examples are intended to demonstrate an approach to the design, and are not intended to suggest that the approach presented is the only approach. The committee responsible for the development of these design examples recognizes that designers have alternate approaches that work best for them and their projects. Design approaches that differ from those presented in these examples are considered viable as long as the AISC Specification, sound engineering, and project specific requirements are satisfied. Part I of these examples is organized to correspond with the organization of the AISC Specification. The Chapter titles match the corresponding chapters in the AISC Specification. Part II is devoted primarily to connection examples that draw on the tables from the AISC Manual, recommended design procedures, and the breadth of the AISC Specification. The chapters of Part II are labeled II-A, II-B, II-C, etc. Part III addresses aspects of design that are linked to the performance of a building as a whole. This includes coverage of lateral stability and second-order analysis, illustrated through a four-story braced-frame and momentframe building. The Design Examples are arranged with LRFD and ASD designs presented side-by-side, for consistency with the AISC Manual. Design with ASD and LRFD are based on the same nominal strength for each element so that the only differences between the approaches are the set of load combinations from ASCE/SEI 7-16 used for design, and whether the resistance factor for LRFD or the safety factor for ASD is used. CONVENTIONS The following conventions are used throughout these examples: 1.

The 2016 AISC Specification for Structural Steel Buildings is referred to as the AISC Specification and the 15th Edition AISC Steel Construction Manual, is referred to as the AISC Manual.

2.

The 2016 ASCE Minimum Design Loads and Associated Criteria for Buildings and Other Structures is referred to as ASCE/SEI 7.

3.

The source of equations or tabulated values taken from the AISC Specification or AISC Manual is noted along the right-hand edge of the page.

4.

When the design process differs between LRFD and ASD, the designs equations are presented side-by-side. This rarely occurs, except when the resistance factor, , and the safety factor, , are applied.

5.

The results of design equations are presented to three significant figures throughout these calculations.

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ACKNOWLEDGMENTS The AISC Committee on Manuals reviewed and approved V15.1 of the AISC Design Examples: Mark V. Holland, Chairman Gary C. Violette, Vice Chairman Allen Adams Scott Adan Abbas Aminmansour Craig Archacki Charles J. Carter Harry A. Cole, Emeritus Brad Davis Bo Dowswell Matt Eatherton Marshall T. Ferrell, Emeritus Patrick J. Fortney Timothy P. Fraser Louis F. Geschwindner, Emeritus John L. Harris III Christopher M. Hewitt William P. Jacobs V Benjamin Kaan

Ronald L. Meng Larry S. Muir Thomas M. Murray James Neary Davis G. Parsons II, Emeritus John Rolfes Rafael Sabelli Thomas J. Schlafly Clifford W. Schwinger William T. Segui, Emeritus Victor Shneur William A. Thornton Michael A. West Ronald G. Yeager Cynthia J. Duncan, Secretary Eric Bolin, Assistant Secretary Michael Gannon, Assistant Secretary Carlo Lini, Assistant Secretary Jennifer Traut-Todaro, Assistant Secretary

The committee gratefully acknowledges the contributions made to this document by the AISC Committee on Specifications and the following individuals: W. Scott Goodrich, Heath Mitchell, William N. Scott, Marc L. Sorenson and Sriramulu Vinnakota.

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TABLE OF CONTENTS PART I

EXAMPLES BASED ON THE AISC SPECIFICATION ........................ I-1

CHAPTER A

GENERAL PROVISIONS ..................................................................................................... A-1

Chapter A References

................................................................................................................................................... A-2

CHAPTER B

DESIGN REQUIREMENTS .................................................................................................. B-1

Chapter B References

................................................................................................................................................... B-2

CHAPTER C

DESIGN FOR STABILITY ................................................................................................... C-1

Example C.1A Example C.1B Example C.1C

Design of a Moment Frame by the Direct Analysis Method ..................................................... C-2 Design of a Moment Frame by the Effective Length Method ................................................... C-7 Design of a Moment Frame by the First-Order Method .......................................................... C-13

CHAPTER D

DESIGN OF MEMBERS FOR TENSION ........................................................................... D-1

Example D.1 Example D.2 Example D.3 Example D.4 Example D.5 Example D.6 Example D.7 Example D.8 Example D.9

W-Shape Tension Member ....................................................................................................... D-2 Single-Angle Tension Member ................................................................................................ D-5 WT-Shape Tension Member .................................................................................................... D-8 Rectangular HSS Tension Member ........................................................................................ D-11 Round HSS Tension Member ................................................................................................. D-14 Double-Angle Tension Member ............................................................................................. D-17 Pin-Connected Tension Member ............................................................................................ D-20 Eyebar Tension Member ........................................................................................................ D-24 Plate with Staggered Bolts ..................................................................................................... D-27

CHAPTER E

DESIGN OF MEMBERS FOR COMPRESSION................................................................ E-1

Example E.1A Example E.1B Example E.1C Example E.1D Example E.2 Example E.3 Example E.4A Example E.4B Example E.5 Example E.6 Example E.7 Example E.8 Example E.9 Example E.10 Example E.11 Example E.12 Example E.13 Example E.14

W-Shape Column Design with Pinned Ends ............................................................................ E-4 W-Shape Column Design with Intermediate Bracing .............................................................. E-6 W-Shape Available Strength Calculation ................................................................................. E-8 W-Shape Available Strength Calculation ............................................................................... E-10 Built-up Column with a Slender Web .................................................................................... E-14 Built-up Column with Slender Flanges .................................................................................. E-19 W-Shape Compression Member (Moment Frame) ................................................................ E-24 W-Shape Compression Member (Moment Frame) ................................................................ E-28 Double-Angle Compression Member without Slender Elements ........................................... E-30 Double-Angle Compression Member with Slender Elements ................................................ E-36 WT Compression Member without Slender Elements ........................................................... E-43 WT Compression Member with Slender Elements ................................................................ E-48 Rectangular HSS Compression Member without Slender Elements ...................................... E-53 Rectangular HSS Compression Member with Slender Elements ........................................... E-56 Pipe Compression Member .................................................................................................... E-61 Built-up I-Shaped Member with Different Flange Sizes ........................................................ E-64 Double-WT Compression Member ......................................................................................... E-70 Eccentrically Loaded Single-Angle Compression Member (Long Leg Attached) .................. E-77

CHAPTER F

DESIGN OF MEMBERS FOR FLEXURE .......................................................................... F-1

Example F.1-1A Example F.1-1B

W-Shape Flexural Member Design in Major Axis Bending, Continuously Braced ................. F-6 W-Shape Flexural Member Design in Major Axis Bending, Continuously Braced .................. F-8

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Example F.1-2A Example F.1-2B Example F.1-3A Example F.1-3B Example F.2-1A Example F.2-1B Example F.2-2A Example F.2-2B Example F.3A Example F.3B Example F.4 Example F.5 Example F.6 Example F.7A Example F.7B Example F.8A Example F.8B Example F.9A Example F.9B Example F.10 Example F.11A Example F.11B Example F.11C Example F.12 Example F.13 Example F.14 Example F.15 Chapter F Design Example References

W-Shape Flexural Member Design in Major Axis Bending, Braced at Third Points ............... F-9 W-Shape Flexural Member Design in Major Axis Bending, Braced at Third Points.............. F-10 W-Shape Flexural Member Design in Major Axis Bending, Braced at Midspan ................... F-12 W-Shape Flexural Member Design in Major Axis Bending, Braced at Midspan ................... F-14 Compact Channel Flexural Member, Continuously Braced .................................................... F-16 Compact Channel Flexural Member, Continuously Braced ................................................... F-18 Compact Channel Flexural Member with Bracing at Ends and Fifth Points .......................... F-19 Compact Channel Flexural Member with Bracing at Ends and Fifth Points .......................... F-20 W-Shape Flexural Member with Noncompact Flanges in Major Axis Bending .................... F-22 W-Shape Flexural Member with Noncompact Flanges in Major Axis Bending .................... F-24 W-Shape Flexural Member, Selection by Moment of Inertia for Major Axis Bending ......... F-26 I-Shaped Flexural Member in Minor Axis Bending .............................................................. .F-28 Square HSS Flexural Member with Compact Flanges ........................................................... F-30 Rectangular HSS Flexural Member with Noncompact Flanges ............................................. F-32 Rectangular HSS Flexural Member with Noncompact Flanges ............................................. F-34 Square HSS Flexural Member with Slender Flanges ............................................................. F-37 Square HSS Flexural Member with Slender Flanges ............................................................. F-39 Pipe Flexural Member ............................................................................................................ F-42 Pipe Flexural Member ............................................................................................................ F-43 WT-Shape Flexural Member .................................................................................................. F-45 Single-Angle Flexural Member with Bracing at Ends Only ................................................... F-48 Single-Angle Flexural Member with Bracing at Ends and Midspan ...................................... F-52 Single Angle Flexural Member with Vertical and Horizontal Loading .................................. F-55 Rectangular Bar in Major Axis Bending ................................................................................ F-62 Round Bar in Bending ............................................................................................................ F-65 Point-Symmetrical Z-shape in Major Axis Bending .............................................................. F-67 Plate Girder Flexural Member ................................................................................................ F-73

CHAPTER G

DESIGN OF MEMBERS FOR SHEAR ...............................................................................G-1

Example G.1A Example G.1B Example G.2A Example G.2B Example G.3 Example G.4 Example G.5 Example G.6 Example G.7 Example G.8A Example G.8B Chapter G Design Example References

W-Shape in Strong Axis Shear ................................................................................................. G-3 W-Shape in Strong Axis Shear ................................................................................................. G-4 Channel in Strong Axis Shear .................................................................................................. G-5 Channel in Strong Axis Shear .................................................................................................. G-6 Angle in Shear .......................................................................................................................... G-8 Rectangular HSS in Shear ...................................................................................................... G-10 Round HSS in Shear ............................................................................................................... G-12 Doubly Symmetric Shape in Weak Axis Shear ...................................................................... G-14 Singly Symmetric Shape in Weak Axis Shear ....................................................................... G-16 Built-up Girder with Transverse Stiffeners ............................................................................ G-18 Built-up Girder with Transverse Stiffeners ............................................................................ G-22

CHAPTER H

DESIGN OF MEMBERS FOR COMBINED FORCES AND TORSION .........................H-1

Example H.1A

W-shape Subject to Combined Compression and Bending About Both Axes (Braced Frame) ............................................................................................ H-2 W-shape Subject to Combined Compression and Bending Moment About Both Axes (Braced Frame) ............................................................................................. H-4 W-Shape Subject to Combined Compression and Bending Moment About Both Axes (By AISC Specification Section H2) ........................................................... H-6 W-Shape Subject to Combined Axial Tension and Flexure ..................................................... H-9

Example H.1B Example H.2 Example H.3

................................................................................................................................................. F-83

................................................................................................................................................. G-25

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Example H.4 Example H.5A Example H.5B Example H.5C Example H.6 Chapter H Design Example References

W-Shape Subject to Combined Axial Compression and Flexure ........................................... H-13 Rectangular HSS Torsional Strength ...................................................................................... H-17 Round HSS Torsional Strength .............................................................................................. H-19 Rectangular HSS Combined Torsional and Flexural Strength ............................................... H-21 W-Shape Torsional Strength .................................................................................................. H-26

CHAPTER I

DESIGN OF COMPOSITE MEMBERS ............................................................................... I-1

Example I.1 Example I.2 Example I.3 Example I.4 Example I.5 Example I.6 Example I.7 Example I.8 Example I.9 Example I.10 Example I.11 Example I.12 Example I.13 Chapter I Design Example References

Composite Beam Design ........................................................................................................... I-4 Composite Girder Design ........................................................................................................ I-15 Filled Composite Member Force Allocation and Load Transfer ............................................. I-34 Filled Composite Member in Axial Compression ................................................................... I-45 Filled Composite Member in Axial Tension ........................................................................... I-50 Filled Composite Member in Combined Axial Compression, Flexure and Shear ................... I-52 Filled Composite Box Column with Noncompact/Slender Elements ...................................... I-66 Encased Composite Member Force Allocation and Load Transfer ......................................... I-82 Encased Composite Member in Axial Compression ............................................................... I-97 Encased Composite Member in Axial Tension ..................................................................... I-104 Encased Composite Member in Combined Axial Compression, Flexure and Shear ............. I-107 Steel Anchors in Composite Components ............................................................................. I-123 Composite Collector Beam Design ....................................................................................... I-127

CHAPTER J

DESIGN OF CONNECTIONS ............................................................................................... J-1

Example J.1 Example J.2 Example J.3 Example J.4A Example J.4B Example J.5 Example J.6

Fillet Weld in Longitudinal Shear ............................................................................................. J-2 Fillet Weld Loaded at an Angle ................................................................................................. J-4 Combined Tension and Shear in Bearing-Type Connections .................................................... J-6 Slip-Critical Connection with Short-Slotted Holes ................................................................... J-8 Slip-Critical Connection with Long-Slotted Holes .................................................................. J-10 Combined Tension and Shear in a Slip-Critical Connection ................................................... J-12 Base Plate Bearing on Concrete ............................................................................................... J-15

CHAPTER K

ADDITIONAL REQUIREMENTS FOR HSS AND BOX-SECTION CONNECTIONS .....................................................................................................................K-1

Example K.1 Example K.2 Example K.3 Example K.4 Example K.5 Example K.6 Example K.7 Example K.8 Example K.9 Example K.10 Chapter K Design Example References

Welded/Bolted Wide Tee Connection to an HSS Column ....................................................... K-2 Welded/Bolted Narrow Tee Connection to an HSS Column ................................................. K-11 Double-Angle Connection to an HSS Column ....................................................................... K-15 Unstiffened Seated Connection to an HSS Column ............................................................... K-19 Stiffened Seated Connection to an HSS Column ................................................................... K-22 Single-Plate Connection to Rectangular HSS Column ........................................................... K-27 Through-Plate Connection to a Rectangular HSS Column .................................................... K-31 Longitudinal Plate Loaded Perpendicular to the HSS Axis on a Round HSS ........................ K-35 Rectangular HSS Column Base Plate ..................................................................................... K-38 Rectangular HSS Strut End Plate ........................................................................................... K-41

APPENDIX 6

MEMBER STABILITY BRACING .................................................................................... A6-1

Example A-6.1

Point Stability Bracing of a W-Shape Column ........................................................................ A6-3

................................................................................................................................................. H-34

................................................................................................................................................ I-136

................................................................................................................................................. K-45

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Example A-6.2 Example A-6.3 Example A-6.4 Example A-6.5 Example A-6.6 Appendix 6 References

Point Stability Bracing of a WT-Shape Column ..................................................................... A6-6 Point Stability Bracing of a BeamCase I ........................................................................... A6-10 Point Stability Bracing of a BeamCase II .......................................................................... A6-14 Point Stability Bracing of a Beam with Reverse Curvature Bending .................................... A6-18 Point Torsional Stability Bracing of a Beam ......................................................................... A6-23

PART II

EXAMPLES BASED ON THE AISC STEEL CONSTRUCTION MANUAL ............................................................................................. II-1

............................................................................................................................................... A6-28

CHAPTER IIA

SIMPLE SHEAR CONNECTIONS ................................................................................ IIA-1

Example II.A-1A Example II.A-1B Example II.A-1C Example II.A-2A Example II.A-2B Example II.A-3 Example II.A-4 Example II.A-5 Example II.A-6 Example II.A-7 Example II.A-8 Example II.A-9 Example II.A-10 Example II.A-11A Example II.A-11B Example II.A-11C Example II.A-12A Example II.A-12B Example II.A-13 Example II.A-14 Example II.A-15 Example II.A-16 Example II.A-17A Example II.A-17B

All-Bolted Double-Angle Connection ............................................................................... IIA-2 All-Bolted Double-Angle Connection Subject to Axial and Shear Loading ...................... IIA-5 All-Bolted Double-Angle Connection—Structural Integrity Check ................................. IIA-24 Bolted/Welded Double-Angle Connection ...................................................................... IIA-31 Bolted/Welded Double-Angle Connection Subject to Axial and Shear Loading ............. IIA-35 All-Welded Double-Angle Connection ........................................................................... IIA-49 All-Bolted Double-Angle Connection in a Coped Beam ................................................. IIA-52 Welded/Bolted Double-Angle Connection in a Coped Beam ........................................... IIA-59 Beam End Coped at the Top Flange Only ....................................................................... IIA-63 Beam End Coped at the Top and Bottom Flanges. .......................................................... IIA-80 All-Bolted Double-Angle Connections (Beams-to-Girder Web) ..................................... IIA-83 Offset All-Bolted Double-Angle Connections (Beams-to-Girder Web) .......................... IIA-96 Skewed Double Bent-Plate Connection (Beam-to-Girder Web). .................................... IIA-99 Shear End-Plate Connection (Beam to Girder Web). .................................................... IIA-105 End-Plate Connection Subject to Axial and Shear Loading ........................................... IIA-107 Shear End-Plate Connection—Structural Integrity Check ............................................. IIA-118 All-Bolted Unstiffened Seated Connection (Beam-to-Column Web) ............................ IIA-124 All-Bolted Unstiffened Seated Connection—Structural Integrity Check ....................... IIA-128 Bolted/Welded Unstiffened Seated Connection (Beam-to-Column Flange) ................. IIA-134 Bolted/Welded Stiffened Seated Connection (Beam-to-Column Flange) ..................... IIA-137 Bolted/Welded Stiffened Seated Connection (Beam-to-Column Web) ......................... IIA-141 Offset Unstiffened Seated Connection (Beam-to-Column Flange). .............................. IIA-145 Single-Plate Connection (Conventional Beam-to-Column Flange) ............................... IIA-148 Single-Plate Connection Subject to Axial and Shear Loading (Beam-to-Column Flange) .............................................................................................. IIA-150 Single-Plate Connection—Structural Integrity Check .................................................... IIA-163 Single-Plate Connection (Beam-to-Girder Web) ........................................................... IIA-169 Extended Single-Plate Connection (Beam-to-Column Web) ......................................... IIA-174 Extended Single-Plate Connection Subject to Axial and Shear Loading ....................... IIA-182 All-Bolted Single-Plate Shear Splice ............................................................................. IIA-205 Bolted/Welded Single-Plate Shear Splice ...................................................................... IIA-211 Bolted Bracket Plate Design .......................................................................................... IIA-217 Welded Bracket Plate Design. ....................................................................................... IIA-224 Eccentrically Loaded Bolt Group (IC Method) ............................................................. IIA-230 Eccentrically Loaded Bolt Group (Elastic Method)....................................................... IIA-232 Eccentrically Loaded Weld Group (IC Method)............................................................ IIA-234 Eccentrically Loaded Weld Group (Elastic Method) ..................................................... IIA-237 All-Bolted Single-Angle Connection (Beam-to-Girder Web) ....................................... IIA-240 All-Bolted Single-Angle Connection—Structural Integrity Check ............................... IIA-250 Bolted/Welded Single-Angle Connection (Beam-to-Column Flange). ......................... IIA-257 All-Bolted Tee Connection (Beam-to-Column Flange) ................................................. IIA-260 Bolted/Welded Tee Connection (Beam-to-Column Flange) .......................................... IIA-270

Example II.A-17C Example II.A-18 Example II.A-19A Example II.A-19B Example II.A-20 Example II.A-21 Example II.A-22 Example II.A-23 Example II.A-24 Example II.A-25 Example II.A-26 Example II.A-27 Example II.A-28A Example II.A-28B Example II.A-29 Example II.A-30 Example II.A-31

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CHAPTER IIB

FULLY RESTRAINED (FR) MOMENT CONNECTIONS ........................................... IIB-1

Example II.B-1 Example II.B-2 Example II.B-3 Chapter IIB Design Example References

Bolted Flange-Plated FR Moment Connection (Beam-to-Column Flange) .......................... IIB-2 Welded Flange-Plated FR Moment Connection (Beam-to-Column Flange) ....................... IIB-20 Directly Welded Flange FR Moment Connection (Beam-to-Column Flange). ................... IIB-27

CHAPTER IIC

BRACING AND TRUSS CONNECTIONS ...................................................................... IIC-1

Example II.C-1 Example II.C-2 Example II.C-3

Truss Support Connection ..................................................................................................... IIC-2 Truss Support Connection ................................................................................................... IIC-16 Heavy Wide Flange Compression Connection (Flanges on the Outside) ............................ IIC-24

CHAPTER IID

MISCELLANEOUS CONNECTIONS .............................................................................. IID-1

Example II.D-1 Example II.D-2 Example II.D-3

WT Hanger Connection ......................................................................................................... IID-2 Beam Bearing Plate ............................................................................................................. IID-10 Slip-Critical Connection with Oversized Holes ................................................................... IID-17

PART III

SYSTEM DESIGN EXAMPLES ......................................................... III-1

Example III-1

Design of Selected Members and Lateral Analysis of a Four-Story Building.......................... III-2 Introduction .............................................................................................................................. III-2 Conventions.............................................................................................................................. III-2 Design Sequence ...................................................................................................................... III-3 General Description of the Building......................................................................................... III-4 Roof Member Design and Selection ........................................................................................ III-6 Select Roof Joists ................................................................................................................ III-7 Select Roof Beams .............................................................................................................. III-8 Select Roof Beams at the End (East & West) of the Building .......................................... III-10 Select Roof Beams at the End (North & South) of the Building....................................... III-13 Select Roof Beams Along the Interior Lines of the Building ........................................... III-17 Floor Member Design and Selection ..................................................................................... III-21 Select Floor Beams (Composite and Noncomposite)........................................................ III-22 Select Typical 45-ft-Long Interior Composite Beam (10 ft on center) ............................. III-22 Select Typical 30-ft Interior Composite (or Noncomposite) Beam (10 ft on center) ........ III-27 Select Typical North-South Edge Beam ........................................................................... III-33 Select Typical East-West Edge Girder .............................................................................. III-36 Select Typical East-West Interior Girder .......................................................................... III-40 Column Design and Selection for Gravity Loads .................................................................. III-46 Select Typical Interior Leaning Columns ......................................................................... III-52 Select Typical Exterior Leaning Columns ........................................................................ III-53 Wind Load Determination ...................................................................................................... III-55 Seismic Load Determination .................................................................................................. III-59 Moment Frame Model ............................................................................................................ III-73 Calculation of Required Strength—Three Methods .............................................................. III-77 Method 1—Direct Analysis Method ................................................................................. III-77 Method 2—Effective Length Method ............................................................................... III-82 Method 3—Simplified Effective Length Method ............................................................. III-87 Beam Analysis in the Moment Frame .................................................................................... III-90 Braced Frame Analysis .......................................................................................................... III-93 Analysis of Drag Struts .......................................................................................................... III-98 Part III Example References................................................................................................. III-101

.............................................................................................................................................. IIB-29

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Part I Examples Based on the AISC Specification This part contains design examples demonstrating select provisions of the AISC Specification for Structural Steel Buildings.

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Chapter A General Provisions A1. SCOPE These design examples are intended to illustrate the application of the 2016 AISC Specification for Structural Steel Buildings, ANSI/AISC 360-16 (AISC, 2016a), and the AISC Steel Construction Manual, 15th Edition (AISC, 2017) in low-seismic applications. For information on design applications requiring seismic detailing, see the 2016 AISC Seismic Provisions for Structural Steel Buildings, ANSI/AISC 341-16 (AISC, 2016b) and the AISC Seismic Design Manual, 2nd Edition (AISC, 2012). A2. REFERENCED SPECIFICATIONS, CODES AND STANDARDS Section A2 includes a detailed list of the specifications, codes and standards referenced throughout the AISC Specification. A3. MATERIAL Section A3 includes a list of the steel materials that are approved for use with the AISC Specification. The complete ASTM standards for the most commonly used steel materials can be found in Selected ASTM Standards for Structural Steel Fabrication (ASTM, 2016). A4. STRUCTURAL DESIGN DRAWINGS AND SPECIFICATIONS Section A4 requires that structural design drawings and specifications meet the requirements in the AISC Code of Standard Practice for Steel Buildings and Bridges, ANSI/AISC 303-16 (AISC, 2016c).

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CHAPTER A REFERENCES AISC (2012), Seismic Design Manual, 2nd Ed., American Institute of Steel Construction, Chicago, IL. AISC (2016a), Specification for Structural Steel Buildings, ANSI/AISC 360-16, American Institute of Steel Construction, Chicago, IL. AISC (2016b), Seismic Provisions for Structural Steel Buildings, ANSI/AISC 341-16, American Institute of Steel Construction, Chicago, IL. AISC (2016c), Code of Standard Practice for Steel Buildings and Bridges, ANSI/AISC 303-16, American Institute of Steel Construction, Chicago, IL. AISC (2017), Steel Construction Manual, 15th Ed., American Institute of Steel Construction, Chicago, IL. ASTM (2016), Selected ASTM Standards for Structural Steel Fabrication, ASTM International, West Conshohocken, PA.

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Chapter B Design Requirements B1. GENERAL PROVISIONS The AISC Specification requires that the design of members and connections shall be consistent with the intended behavior of the framing system and the assumptions made in the structural analysis. B2. LOADS AND LOAD COMBINATIONS In the absence of an applicable building code, the default load combinations to be used with the AISC Specification are those from Minimum Design Loads and Associated Criteria for Buildings and Other Structures, ASCE/SEI 7-16 (ASCE, 2016). B3. DESIGN BASIS Chapter B of the AISC Specification and Part 2 of the AISC Manual describe the basis of design, for both load and resistance factor design (LRFD) and allowable strength design (ASD). AISC Specification Section B3.4 describes three basic types of connections: simple connections, fully restrained (FR) moment connections, and partially restrained (PR) moment connections. Several examples of the design of each of these types of connections are given in Part II of these Design Examples. Information on the application of serviceability and ponding provisions may be found in AISC Specification Chapter L and AISC Specification Appendix 2, respectively, and their associated commentaries. Design examples and other useful information on this topic are given in AISC Design Guide 3, Serviceability Design Considerations for Steel Buildings, Second Edition (West et al., 2003). Information on the application of fire design provisions may be found in AISC Specification Appendix 4 and its associated commentary. Design examples and other useful information on this topic are presented in AISC Design Guide 19, Fire Resistance of Structural Steel Framing (Ruddy et al., 2003). Corrosion protection and fastener compatibility are discussed in Part 2 of the AISC Manual. B4. MEMBER PROPERTIES AISC Specification Tables B4.1a and B4.1b give the complete list of limiting width-to-thickness ratios for all compression and flexural members defined by the AISC Specification. Except for one section, the W-shapes presented in the compression member selection tables as column sections meet the criteria as nonslender element sections. The W-shapes with a nominal depth of 8 in. or larger presented in the flexural member selection tables as beam sections meet the criteria for compact sections, except for seven specific shapes. When noncompact or slender-element sections are tabulated in the design aids, local buckling criteria are accounted for in the tabulated design values. The shapes listing and other member design tables in the AISC Manual also include footnoting to highlight sections that exceed local buckling limits in their most commonly available material grades. These footnotes include the following notations for W-shapes: c

Shape is slender for compression with Fy = 50 ksi. Shape exceeds compact limit for flexure with Fy = 50 ksi. g The actual size, combination and orientation of fastener components should be compared with the geometry of the cross section to ensure compatibility. f

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h v

Flange thickness greater than 2 in. Special requirements may apply per AISC Specification Section A3.1c. Shape does not meet the h/tw limit for shear in AISC Specification Section G2.1(a) with Fy = 50 ksi.

CHAPTER B REFERENCES ASCE (2016), Minimum Design Loads and Associated Criteria for Buildings and Other Structures, ASCE/SEI 716, American Society of Civil Engineers, Reston, VA. West, M.A., Fisher, J.M. and Griffis, L.G. (2003), Serviceability Design Considerations for Steel Buildings, Design Guide 3, 2nd Ed., AISC, Chicago, IL. Ruddy, J.L., Marlo, J.P., Ioannides, S.A. and Alfawakhiri, F. (2003), Fire Resistance of Structural Steel Framing, Design Guide 19, AISC, Chicago, IL.

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Chapter C Design for Stability C1. GENERAL STABILITY REQUIREMENTS The AISC Specification requires that the designer account for both the stability of the structural system as a whole and the stability of individual elements. Thus, the lateral analysis used to assess stability must include consideration of the combined effect of gravity and lateral loads, as well as member inelasticity, out-of-plumbness, out-ofstraightness, and the resulting second-order effects, P-and P-. The effects of “leaning columns” must also be considered, as illustrated in the examples in this chapter and in the four-story building design example in Part III of these Design Examples. P-and P- effects are illustrated in AISC Specification Commentary Figure C-C2.1. Methods for addressing stability, including P-and P- effects, are provided in AISC Specification Section C2 and Appendix 7. C2. CALCULATION OF REQUIRED STRENGTHS The calculation of required strengths is illustrated in the examples in this chapter and in the four-story building design example in Part III of these Design Examples. C3. CALCULATION OF AVAILABLE STRENGTHS The calculation of available strengths is illustrated in the four-story building design example in Part III of these Design Examples.

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EXAMPLE C.1A DESIGN OF A MOMENT FRAME BY THE DIRECT ANALYSIS METHOD Given: Determine the required strengths and effective length factors for the columns in the moment frame shown in Figure C.1A-1 for the maximum gravity load combination, using LRFD and ASD. The uniform load, wD, includes beam self-weight and an allowance for column self-weight. Use the direct analysis method. All members are ASTM A992 material. Columns are unbraced between the footings and roof in the x- and y-axes and have pinned bases.

Fig. C.1A-1. Example C.1A moment frame elevation. Solution: From AISC Manual Table 1-1, the W1265 has A = 19.1 in.2 The beams from grid lines A to B and C to E and the columns at A, D and E are pinned at both ends and do not contribute to the lateral stability of the frame. There are no P- effects to consider in these members and they may be designed using Lc  L. The moment frame between grid lines B and C is the source of lateral stability and therefore will be evaluated using the provisions of Chapter C of the AISC Specification. Although the columns at grid lines A, D and E do not contribute to lateral stability, the forces required to stabilize them must be considered in the moment-frame analysis. The entire frame from grid line A to E could be modeled, but in this case the model is simplified as shown in Figure C.1A-2, in which the stability loads from the three “leaning” columns are combined into a single representative column. From Chapter 2 of ASCE/SEI 7, the maximum gravity load combinations are: LRFD

ASD wu  D  L

wu  1.2 D  1.6 L  1.2  0.400 kip/ft   1.6 1.20 kip/ft   2.40 kip/ft

 0.400 kip/ft  1.20 kip/ft  1.60 kip/ft

Per AISC Specification Section C2.1(d), for LRFD, perform a second-order analysis and member strength checks using the LRFD load combinations. For ASD, perform a second-order analysis using 1.6 times the ASD load combinations and divide the analysis results by 1.6 for the ASD member strength checks.

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Frame analysis gravity loads The uniform gravity loads to be considered in a second-order analysis on the beam from B to C are: wu  2.40 kip/ft

LRFD

wa  1.6 1.60 kip/ft 

ASD

 2.56 kip/ft

Concentrated gravity loads to be considered in a second-order analysis on the columns at B and C contributed by adjacent beams are: LRFD wu l Pu  2  2.40 kip/ft  30.0 ft   2  36.0 kips

ASD wa l Pa  2  2.56 kip/ft  30.0 ft   2  38.4 kips

Concentrated gravity loads on the representative “leaning” column The load in this column accounts for all gravity loading that is stabilized by the moment frame, but is not directly applied to it. LRFD    60.0 ft  2.40 kip/ft  PuL

ASD    60.0 ft  2.56 kip/ft  PaL

 144 kips

 154 kips

Frame analysis notional loads Per AISC Specification Section C2.2, frame out-of-plumbness must be accounted for either by explicit modeling of the assumed out-of-plumbness or by the application of notional loads. Use notional loads. From AISC Specification Equation C2-1, the notional loads are: LRFD

ASD

  1.0

  1.6

Yi  120 ft  2.40 kip ft 

Yi  120 ft 1.60 kip ft 

 288 kips Ni  0.002Yi



 Spec. Eq. C2-1

 192 kips Ni  0.002Yi

 0.002 1.0  288 kips 

 0.002 1.6 192 kips 

 0.576 kip

 0.614 kip

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 Spec. Eq. C2-1

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Summary of applied frame loads The applied loads are shown in Figure C.1A-2. LRFD

ASD

Fig. C.1A-2. Applied loads on the analysis model. Per AISC Specification Section C2.3, conduct the analysis using 80% of the nominal stiffnesses to account for the effects of inelasticity. Assume, subject to verification, that Pr /Pns is not greater than 0.5; therefore, no additional stiffness reduction is required (b = 1.0). Half of the gravity load is carried by the columns of the moment-resisting frame. Because the gravity load supported by the moment-resisting frame columns exceeds one-third of the total gravity load tributary to the frame, per AISC Specification Section C2.1, the effects of P- and P-must be considered in the frame analysis. This example uses analysis software that accounts for both P- and P- effects. (If the software used does not account for P- effects this may be accomplished by subdividing the columns between the footing and beam.) Figures C.1A-3 and C.1A-4 show results from a first-order and a second-order analysis. (The first-order analysis is shown for reference only.) In each case, the drift is the average of drifts at grid lines B and C. First-order results LRFD 1st  0.181 in. 

1st

ASD (Reactions and moments divided by 1.6)  0.193 in. (prior to dividing by 1.6)

Fig. C.1A-3. Results of first-order analysis.

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Second-order results LRFD

 2nd  0.290 in. 

 2 nd

ASD (Reactions and moments divided by 1.6)  0.321 in. (prior to dividing by 1.6) 

 Drift ratio:   2nd 0.321 in.  1st 0.193 in.   1.66

Drift ratio:

 2nd 0.290 in.  1st 0.181 in.   1.60

Fig. C.1A-4. Results of second-order analysis.

Check the assumption that Pr Pns  0.5 on the column on grid line C. Because a W1265 column contains no elements that are slender for uniform compression, Pns  Fy Ag



  50 ksi  19.1 in.2



 955 kips

Pr 1.0  72.6 kips   Pns 955kips

LRFD

 0.0760  0.5 o.k.

Pr 1.6  48.4 kips   Pns 955kips

ASD

 0.0811  0.5 o.k.

The stiffness assumption used in the analysis, b = 1.0, is verified. Note that the drift ratio, 1.60 (LRFD) or 1.66 (ASD), does not exceed the recommended limit of 2.5 from AISC Specification Commentary Section C1. The required axial compressive strength in the columns is 72.6 kips (LRFD) or 48.4 kips (ASD). The required bending moment diagram is linear, varying from zero at the bottom to 127 kip-ft (LRFD) or 84.8 kip-ft (ASD) at the top. These required strengths apply to both columns because the notional load must be applied in each direction.

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Although the second-order sway multiplier (drift ratio) is fairly large at 1.60 (LRFD) or 1.66 (ASD), the change in bending moment is small because the only sway moments are those produced by the small notional loads. For load combinations with significant gravity and lateral loadings, the increase in bending moments is larger. Per AISC Specification Section C3, the effective length for flexural buckling of all members is taken as the unbraced length (K = 1.0): Lcx  20.0 ft Lcy  20.0 ft

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EXAMPLE C.1B DESIGN OF A MOMENT FRAME BY THE EFFECTIVE LENGTH METHOD Given:

Repeat Example C.1A using the effective length method. Determine the required strengths and effective length factors for the columns in the moment frame shown in Figure C.1B-1 for the maximum gravity load combination, using LRFD and ASD. Use the effective length method. Columns are unbraced between the footings and roof in the x- and y-axes and have pinned bases.

Fig. C.1B-1. Example C.1B moment frame elevation. Solution:

From AISC Manual Table 1-1, the W1265 has Ix = 533 in.4 The beams from grid lines A to B and C to E and the columns at A, D and E are pinned at both ends and do not contribute to the lateral stability of the frame. There are no P- effects to consider in these members and they may be designed using Lc  L. The moment frame between grid lines B and C is the source of lateral stability and therefore will be evaluated using the provisions of Chapter C of the AISC Specification. Although the columns at grid lines A, D and E do not contribute to lateral stability, the forces required to stabilize them must be considered in the moment-frame analysis. The entire frame from grid line A to E could be modeled, but in this case the model is simplified as shown in Figure C.1B-2, in which the stability loads from the three “leaning” columns are combined into a single representative column. Check the limitations for the use of the effective length method given in AISC Specification Appendix 7, Section 7.2.1: (a) The structure supports gravity loads primarily through nominally vertical columns, walls or frames. (b) The ratio of maximum second-order drift to the maximum first-order drift (both determined for LRFD load combinations or 1.6 times ASD load combinations, with stiffness not adjusted as specified in AISC Specification Section C2.3) in all stories will be assumed to be no greater than 1.5, subject to verification in the following.

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From Chapter 2 of ASCE/SEI 7, the maximum gravity load combinations are: LRFD

ASD wu  D  L

wu  1.2 D  1.6 L  1.2  0.400 kip/ft   1.6 1.20 kip/ft   2.40 kip/ft

 0.400 kip/ft  1.20 kip/ft  1.60 kip/ft

Per AISC Specification Appendix 7, Section 7.2.2, the analysis must conform to the requirements of AISC Specification Section C2.1, with the exception of the stiffness reduction required by the provisions of Section C2.1(a). Per AISC Specification Section C2.1(d), for LRFD perform a second-order analysis and member strength checks using the LRFD load combinations. For ASD, perform a second-order analysis at 1.6 times the ASD load combinations and divide the analysis results by 1.6 for the ASD member strength checks. Frame analysis gravity loads

The uniform gravity loads to be considered in a second-order analysis on the beam from B to C are: wu  2.40 kip/ft

LRFD

wa  1.6 1.60 kip/ft 

ASD

 2.56 kip/ft

Concentrated gravity loads to be considered in a second-order analysis on the columns at B and C contributed by adjacent beams are: LRFD wu l Pu  2  2.40 kip/ft  30.0 ft   2  36.0 kips

ASD wa l Pa  2  2.56 kip/ft  30.0 ft   2  38.4 kips

Concentrated gravity loads on the representative “leaning” column

The load in this column accounts for all gravity loads that is stabilized by the moment frame, but not directly applied to it. LRFD    60.0 ft  2.40 kip/ft  PuL  144 kips

ASD    60.0 ft  2.56 kip/ft  PaL  154 kips

Frame analysis notional loads

Per AISC Specification Appendix 7, Section 7.2.2, frame out-of-plumbness must be accounted for by the application of notional loads in accordance with AISC Specification Section C2.2b. Note that notional loads need to only be applied to the gravity load combinations per AISC Specification Section C2.2b(d) when the requirement that  2 nd / 1st  1.7 (using stiffness adjusted as specified in Section C2.3) is satisfied. Per the User Note in AISC Specification Appendix 7, Section 7.2.2, Section C2.2b(d) will be satisfied in all cases where the effective length method is applicable, and therefore the notional load need only be applied in gravity-only load cases. From AISC Specification Equation C2-1, the notional loads are:

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LRFD

ASD

  1.0

  1.6

Yi  120 ft  2.40 kip ft 

Yi  120 ft 1.60 kip ft 

 288 kips

 192 kips

 Spec. Eq. C2-1

Ni  0.002Yi

 Spec. Eq. C2-1

Ni  0.002Yi

 0.002 1.0  288 kips 

 0.002 1.6 192 kips 

 0.576 kip

 0.614 kip

Summary of applied frame loads

The applied loads are shown in Figure C.1B-2. LRFD

ASD

Fig. C.1B-2. Applied loads on the analysis model.

Per AISC Specification Appendix 7, Section 7.2.2, conduct the analysis using the full nominal stiffnesses. Half of the gravity load is carried by the columns of the moment-resisting frame. Because the gravity load supported by the moment-resisting frame columns exceeds one-third of the total gravity load tributary to the frame, per AISC Specification Section C2.1(b), the effects of P- on the response of the structure must be considered in the frame analysis. This example uses analysis software that accounts for both P- and P- effects. When using software that does not account for P- effects, this could be accomplished by subdividing columns between the footing and beam. Figures C.1B-3 and C.1B-4 show results from a first-order and second-order analysis. In each case, the drift is the average of drifts at grid lines B and C.

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First-order results

LRFD 1st = 0.145 in.

ASD (Reactions and moments divided by 1.6) 1st = 0.155 in. (prior to dividing by 1.6)

Fig. C.1B-3. Results of first-order analysis. Second-order results

LRFD

ASD

 2nd  0.204 in.

2nd  0.223 in. (prior to dividing by 1.6)

Drift ratio:

Drift ratio:

2nd 0.204 in.  1st 0.145 in.  1.41

2nd 0.223 in.  1st 0.155 in.  1.44

Fig. C-1B-4. Results of second-order analysis.

The assumption that the ratio of the maximum second-order drift to the maximum first-order drift is no greater than 1.5 is verified; therefore, the effective length method is permitted. Although the second-order sway multiplier is fairly large at approximately 1.41 (LRFD) or 1.44 (ASD), the change in bending moment is small because the only sway moments for this load combination are those produced by the small notional loads. For load combinations with significant gravity and lateral loadings, the increase in bending moments is larger. Calculate the in-plane effective length factor, Kx, using the “story stiffness approach” and Equation C-A-7-5 presented in AISC Specification Commentary Appendix 7, Section 7.2. With Kx = K2:

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Pstory RM Pr

Kx 

 2 EI  2  L

  H       HL 

2 EI   H    L2  1.7 H col L 

(Spec. Eq. C-A-7-5)

Calculate the total load in all columns, Pstory , as follows: LRFD Pstory   2.40 kip/ft 120 ft 

ASD Pstory  1.60 kip/ft 120 ft 

 288 kips

 192 kips

Calculate the coefficient to account for the influence of P- on P-, RM, as follows, using AISC Specification Commentary Appendix 7, Equation C-A-7-6: LRFD Pmf  71.5 kips  72.5 kips

ASD Pmf  47.6 kips  48.4 kips  96.0 kips

 144 kips RM  1  0.15  Pmf Pstory 

(Spec. Eq. C-A-7-6)

RM  1  0.15  Pmf Pstory   96.0 kips   1  0.15    192 kips   0.925

 144 kips   1  0.15    288 kips   0.925

Calculate the Euler buckling strength of one moment frame.  2 EI 2

L





 2  29, 000 ksi  533 in.4

 20.0 ft 12 in./ft    2, 650 kips



2

From AISC Specification Commentary Equation C-A-7-5, for the column at line C:

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(Spec. Eq. C-A-7-6)

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LRFD Kx 

Pstory RM Pr

  2 EI   2  L



  EI  2  L 2

ASD

  H       HL 

Kx 

 H     1.7 H L col   

  2 EI   2  L

  288 kips    2, 650 kips    0.925  72.5 kips     0.145 in.     0.576 kip  20.0 ft 12 in./ft  



 2, 650 kips  

Use Kx = 3.45

  EI  2  L

  H       HL 

  H      1.7 1.6  H col L 

  1.6 192 kips     2, 650 kips   0.925 1.6  48.4 kips     0.155 in.     0.614 kip  20.0 ft 12 in./ft  

 2, 650 kips 

  0.145 in.   1.7 6.21 kips 20.0 ft 12 in./ft      

 3.45  0.389

1.6 Pstory RM 1.6  Pr

2



  0.155 in.   4.14 kips 20.0 ft 12 in./ft 1.7 1.6          3.46  0.390

Use Kx = 3.46

Note that the column loads are multiplied by 1.6 for ASD in Equation C-A-7-5. With Kx = 3.45 and Ky = 1.00, the column available strengths can be verified for the given member sizes for the second-order forces (calculations not shown), using the following effective lengths:

Lcx  K x Lx  3.45  20.0 ft   69.0 ft Lcy  K y Ly  1.00  20.0 ft   20.0 ft

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EXAMPLE C.1C DESIGN OF A MOMENT FRAME BY THE FIRST-ORDER METHOD Given:

Repeat Example C.1A using the first-order analysis method. Determine the required strengths and effective length factors for the columns in the moment frame shown in Figure C.1C-1 for the maximum gravity load combination, using LRFD and ASD. Use the first-order analysis method as given in AISC Specification Appendix 7, Section 7.3. Columns are unbraced between the footings and roof in the x- and y-axes and have pinned bases.

Fig. C.1C-1. Example C.1C moment frame elevation. Solution:

From AISC Manual Table 1-1, the W1265 has A = 19.1 in.2 The beams from grid lines A to B and C to E and the columns at A, D and E are pinned at both ends and do not contribute to the lateral stability of the frame. There are no P- effects to consider in these members and they may be designed using Lc=L. The moment frame between grid lines B and C is the source of lateral stability and will be designed using the provisions of AISC Specification Appendix 7, Section 7.3. Although the columns at grid lines A, D and E do not contribute to lateral stability, the forces required to stabilize them must be considered in the moment-frame analysis. These members need not be included in the analysis model, except that the forces in the “leaning” columns must be included in the calculation of notional loads. Check the limitations for the use of the first-order analysis method given in AISC Specification Appendix 7, Section 7.3.1: (a) The structure supports gravity loads primarily through nominally vertical columns, walls or frames. (b) The ratio of maximum second-order drift to the maximum first-order drift (both determined for LRFD load combinations or 1.6 times ASD load combinations, with stiffnesses not adjusted as specified in AISC Specification Section C2.3) in all stories will be assumed to be equal to or less than 1.5, subject to verification. (c) The required axial compressive strength of all members whose flexural stiffnesses are considered to contribute to the lateral stability of the structure will be assumed to be no more than 50% of the crosssection strength, subject to verification. Per AISC Specification Appendix 7, Section 7.3.2, the required strengths are determined from a first-order analysis using notional loads determined in the following, along with a B1 multiplier to account for second-order effects, as determined from Appendix 8.

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Loads From Chapter 2 of ASCE/SEI 7, the maximum gravity load combinations are: LRFD

ASD wu  D  L

wu  1.2 D  1.6 L  1.2  0.400 kip/ft   1.6 1.20 kip/ft   2.40 kip/ft

 0.400 kip/ft  1.20 kip/ft  1.60 kip/ft

Concentrated gravity loads to be considered on the columns at B and C contributed by adjacent beams are: LRFD wu l Pu  2  2.40 kip/ft  30.0 ft   2  36.0 kips

ASD wa l Pa  2 1.60 kip/ft  30.0 ft   2  24.0 kips

Using AISC Specification Appendix 7, Section 7.3.2, frame out-of-plumbness is accounted for by the application of an additional lateral load. From AISC Specification Appendix Equation A-7-2, the additional lateral load is determined as follows:

  1.0 

LRFD

  1.6

ASD

Yi  120 ft 1.60 kip/ft 

Yi  120 ft  2.40 kip/ft 

 192 kips

 288 kips

 = 0 in. (no drift for this load combination)

 = 0 in. (no drift for this load combination)

L   20.0 ft 12 in./ft 

L   20.0 ft 12 in./ft   240 in.

 240 in.

N i  2.1   L  Yi  0.0042Yi

(Spec. Eq. A-7-2)

N i  2.1   L  Yi  0.0042Yi

 0 in.   2.11.0     288 kips   240 in.   0.0042  288 kips 

 0 in.   2.11.6    192 kips   240 in.   0.0042 192 kips 

 0 kip  1.21 kips

 0 kip  0.806 kip

Use Ni = 1.21 kips

Use Ni = 0.806 kip

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(Spec. Eq. A-7-2)

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Summary of applied frame loads The applied loads are shown in Figure C.1C-2. LRFD

ASD

Fig. C.1C-2. Applied loads on the analysis model. Conduct the analysis using the full nominal stiffnesses, as indicated in AISC Specification Commentary Appendix 7, Section 7.3. Using analysis software, the first-order results shown in Figure C.1C-3 are obtained: LRFD

1st  0.203 in. 

1st  0.304 in.

ASD

Fig. C.1C-3. Results of first-order analysis. Check the assumption that the ratio of the second-order drift to the first-order drift does not exceed 1.5. B2 can be used to check this limit. Calculate B2 per Appendix 8, Section 8.2.2 using the results of the first-order analysis. Pmf

LRFD  2  36.0 kips    30.0 ft  2.40 kip/ft   144 kips

Pstory  144 kips  4  36.0 kips   288 kips

Pmf

ASD  2  24.0 kips    30.0 ft 1.60 kip/ft   96.0 kips

Pstory  96.0 kips  4  24.0 kips   192 kips

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LRFD RM  1  0.15  Pmf Pstory 

(Spec. Eq. A-8-8)

ASD RM  1  0.15  Pmf Pstory 

 1  0.15 144 kips 288 kips 

 1  0.15  96.0 kips 192 kips 

 0.925

 0.925

 H  0.304 in.

 H  0.203 in.

H  6.53 kips  5.32 kips

H  4.35 kips  3.55 kips  0.800 kip

= 1.21 kips L   20 ft 12 in./ft 

L   20 ft 12 in./ft 

 240 in.

 240 in.

HL H (1.21 kips)  240 in.   0.925 0.304 in.  884 kips

Pe story  RM

(Spec. Eq. A-8-7)

 = 1.0 B2 

(Spec. Eq. A-8-8)

HL H  0.800 kip  240 in.  0.925 0.203 in.  875 kips

Pe story  RM

(Spec. Eq. A-8-7)

 = 1.6

1 1 Pstory 1 Pe story

(Spec. Eq. A-8-6)

1 1 1.0  288 kips  1 884 kips  1.48  1

B2 

1 1 Pstory 1 Pe story

(Spec. Eq. A-8-6)

1 1 1.6 192 kips  1 875 kips  1.54  1





When a structure with a live-to-dead load ratio of 3 is analyzed by a first-order analysis the required strength for LRFD will always be 1.5 times the required strength for ASD. However, when a second-order analysis is used this ratio is not maintained. This is due to the use of the amplification factor,, which is set equal to 1.6 for ASD, in order to capture the worst case second-order effects for any live-to-dead load ratio. Thus, in this example the limitation for applying the first-order analysis method, that the ratio of the maximum second-order drift to maximum first-order drift is not greater than 1.5, is verified for LRFD but is not verified for ASD. Therefore, for this example the first-order method is invalid for ASD and will proceed with LRFD only. Check the assumption that  Pr  0.5 Pns and, therefore, the first-order analysis method is permitted.

Because the W1265 column does not contain elements that are slender for compression, Pns  Fy Ag 0.5 Pns  0.5 Fy Ag



 0.5  50 ksi  19.1 in.2



 478 kips

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Pr  1.0  72.8 kips   72.8 kips  478 kips o.k. (LRFD only)

The assumption that the first-order analysis method can be used is verified for LRFD. Although the second-order sway multiplier is 1.48, the change in bending moment is small because the only sway moments are those produced by the small notional loads. For load combinations with significant gravity and lateral loadings, the increase in bending moments is larger. The column strengths can be verified after using the B1 amplification given in Appendix 8, Section 8.2.1 to account for second-order effects (calculations not shown here). In the direction of sway, the effective length factor is taken equal to 1.00, and the column effective lengths are as follows: Lcx  20.0 ft Lcy  20.0 ft

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Chapter D Design of Members for Tension D1. SLENDERNESS LIMITATIONS AISC Specification Section D1 does not establish a slenderness limit for tension members, but recommends limiting L/r to a maximum of 300. This is not an absolute requirement. Rods and hangers are specifically excluded from this recommendation. D2. TENSILE STRENGTH Both tensile yielding strength and tensile rupture strength must be considered for the design of tension members. It is not unusual for tensile rupture strength to govern the design of a tension member, particularly for small members with holes or heavier sections with multiple rows of holes. For preliminary design, tables are provided in Part 5 of the AISC Manual for W-shapes, L-shapes, WT-shapes, rectangular HSS, square HSS, round HSS, Pipe, and 2L-shapes. The calculations in these tables for available tensile rupture strength assume an effective area, Ae, of 0.75Ag. The gross area, Ag, is the total cross-sectional area of the member. If the actual effective area is greater than 0.75Ag, the tabulated values will be conservative and calculations can be performed to obtain higher available strengths. If the actual effective area is less than 0.75Ag, the tabulated values will be unconservative and calculations are necessary to determine the available strength. D3. EFFECTIVE NET AREA In computing net area, An, AISC Specification Section B4.3b requires that an extra z in. be added to the bolt hole diameter. A computation of the effective area for a chain of holes is presented in Example D.9. Unless all elements of the cross section are connected, Ae  AnU , where U is a reduction factor to account for shear lag. The appropriate values of U can be obtained from AISC Specification Table D3.1. D4. BUILT-UP MEMBERS The limitations for connections of built-up members are discussed in Section D4 of the AISC Specification. D5. PIN-CONNECTED MEMBERS An example of a pin-connected member is given in Example D.7. D6. EYEBARS An example of an eyebar is given in Example D.8. The strength of an eyebar meeting the dimensional requirements of AISC Specification Section D6 is governed by tensile yielding of the body.

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EXAMPLE D.1

W-SHAPE TENSION MEMBER

Given: Select an ASTM A992 W-shape with 8 in. nominal depth to carry a dead load of 30 kips and a live load of 90 kips in tension. The member is 25.0 ft long. Verify the member strength by both LRFD and ASD with the bolted end connection as shown in Figure D.1-1. Verify that the member satisfies the recommended slenderness limit. Assume that connection limit states do not govern.

Fig D.1-1. Connection geometry for Example D.1. Solution: From Chapter 2 of ASCE/SEI 7, the required tensile strength is: LRFD Pu  1.2  30 kips   1.6  90 kips 

ASD Pa  30 kips  90 kips  120 kips

 180 kips From AISC Manual Table 5-1, try a W821.

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W821

Ag bf tf d ry

= 6.16 in.2 = 5.27 in. = 0.400 in. = 8.28 in. = 1.26 in.

The WT-shape corresponding to a W821 is a WT410.5. From AISC Manual Table 1-8, the geometric properties are as follows: WT410.5 y = 0.831 in.

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Tensile Yielding From AISC Manual Table 5-1, the available tensile yielding strength of a W821 is:

ASD

LRFD t Pn  277 kips  180 kips

Pn  184 kips  120 kips t

o.k.

o.k.

Tensile Rupture Verify the table assumption that Ae Ag  0.75 for this connection. From the description of the element in AISC Specification Table D3.1, Case 7, calculate the shear lag factor, U, as the larger of the values from AISC Specification Section D3, Table D3.1 Case 2 and Case 7. From AISC Specification Section D3, for open cross sections, U need not be less than the ratio of the gross area of the connected element(s) to the member gross area. U 

2b f t f Ag 2  5.27 in. 0.400 in. 6.16 in.2

 0.684

Case 2: Determine U based on two WT-shapes per AISC Specification Commentary Figure C-D3.1, with x  y  0.831 in. and where l is the length of connection. x l 0.831 in.  1 9.00 in.  0.908

U  1

Case 7: b f  5.27 in. 2 2 d   8.28 in. 3 3  5.52 in.

Because the flange is connected with three or more fasteners per line in the direction of loading and b f  U = 0.85 Therefore, use the larger U = 0.908. Calculate An using AISC Specification Section B4.3b.

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2 d: 3

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An  Ag  4  d h  z in. t f  6.16 in.2  4 m in.  z in. 0.400 in.  4.76 in.2

Calculate Ae using AISC Specification Section D3. Ae  AnU



2

 4.76 in.

(Spec. Eq. D3-1)

  0.908

 4.32 in.2

Ae 4.32 in.2  Ag 6.16 in.2  0.701  0.75

Because Ae/Ag < 0.75, the tensile rupture strength from AISC Manual Table 5-1 is not valid. The available tensile rupture strength is determined using AISC Specification Section D2 as follows: Pn  Fu Ae



  65 ksi  4.32 in.

2

(Spec. Eq. D2-2)



 281 kips

From AISC Specification Section D2, the available tensile rupture strength is: LRFD

ASD

t = 0.75  t Pn  0.75  281 kips 

t = 2.00 Pn 281 kips  t 2.00  141 kips  120 kips

 211 kips  180 kips o.k.

o.k.

Note that the W821 available tensile strength is governed by the tensile rupture limit state at the end connection versus the tensile yielding limit state. See Chapter J for illustrations of connection limit state checks. Check Recommended Slenderness Limit L  25.0 ft 12 in./ft   r 1.26 in.  238  300 from AISC Specification Section D1 o.k.

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EXAMPLE D.2

SINGLE-ANGLE TENSION MEMBER

Given: Verify the tensile strength of an ASTM A36 L442 with one line of four w-in.-diameter bolts in standard holes, as shown in Figure D.2-1. The member carries a dead load of 20 kips and a live load of 60 kips in tension. Additionally, calculate at what length this tension member would cease to satisfy the recommended slenderness limit. Assume that connection limit states do not govern.

Fig. D.2-1. Connection geometry for Example D.2.

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-7, the geometric properties are as follows: L442

Ag = 3.75 in.2 rz = 0.776 in. x = 1.18 in.

From Chapter 2 of ASCE/SEI 7, the required tensile strength is: LRFD Pu  1.2  20 kips   1.6  60 kips   120 kips

ASD Pa  20 kips  60 kips  80.0 kips

Tensile Yielding

Pn  Fy Ag

(Spec. Eq. D2-1)



  36 ksi  3.75 in.2



 135 kips From AISC Specification Section D2, the available tensile yielding strength is:

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LRFD

ASD

t  0.90 

 t  1.67 

t Pn  0.90 135 kips 

Pn 135 kips  t 1.67  80.8 kips  80.0 kips o.k.

 122 kips  120 kips o.k.

Tensile Rupture From the description of the element in AISC Specification Table D3.1 Case 8, calculate the shear lag factor, U, as the larger of the values from AISC Specification Section D3, Table D3.1 Case 2 and Case 8. From AISC Specification Section D3, for open cross sections, U need not be less than the ratio of the gross area of the connected element(s) to the member gross area. Half of the member is connected, therefore, the minimum value of U is: U = 0.500 Case 2, where l is the length of connection and y  x : x l 1.18 in.  1 9.00 in.  0.869

U  1

Case 8, with four or more fasteners per line in the direction of loading: U = 0.80 Therefore, use the larger U = 0.869. Calculate An using AISC Specification Section B4.3b. An  Ag   d h  z in. t  3.75 in.  m in.  z in.2 in.  3.31 in.2

Calculate Ae using AISC Specification Section D3. Ae  AnU



2

 3.31 in.

(Spec. Eq. D3-1)

  0.869 

 2.88 in.2 Pn  Fu Ae



  58 ksi  2.88 in.2

(Spec. Eq. D2-2)



 167 kips

From AISC Specification Section D2, the available tensile rupture strength is:

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LRFD t  0.75 

 t  2.00 

 t Pn  0.75 167 kips 



 125 kips  120 kips o.k.

ASD

Pn 167 kips  t 2.00  83.5 kips  80.0 kips o.k.



The L442 available tensile strength is governed by the tensile yielding limit state. ASD

LRFD t Pn  122 kips  120 kips

Pn  80.8 kips  80.0 kips o.k. t

o.k.

Recommended Lmax Using AISC Specification Section D1: Lmax  300rz  0.776 in.   300    12 in./ft   19.4 ft Note: The L/r limit is a recommendation, not a requirement. See Chapter J for illustrations of connection limit state checks.

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EXAMPLE D.3

WT-SHAPE TENSION MEMBER

Given:

An ASTM A992 WT620 member has a length of 30 ft and carries a dead load of 40 kips and a live load of 120 kips in tension. As shown in Figure D3-1, the end connection is fillet welded on each side for 16 in. Verify the member tensile strength by both LRFD and ASD. Assume that the gusset plate and the weld are satisfactory.

Fig. D.3-1. Connection geometry for Example D.3.

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-8, the geometric properties are as follows: WT620

= 5.84 in.2 = 8.01 in. = 0.515 in. = 1.57 in. y = 1.09 in.

Ag bf tf rx

From Chapter 2 of ASCE/SEI 7, the required tensile strength is: LRFD Pu  1.2  40 kips   1.6 120 kips 

ASD Pa  40 kips  120 kips  160 kips

 240 kips Tensile Yielding

Check tensile yielding limit state using AISC Manual Table 5-3. LRFD t Pn  263 kips  240 kips

o.k. 

ASD Pn  175 kips  160 kips o.k. t

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Tensile Rupture Check tensile rupture limit state using AISC Manual Table 5-3. LRFD t Pn  214 kips  240 kips

ASD Pn  142 kips  160 kips t

n.g. 

n.g.

The tabulated available rupture strengths don’t work and may be conservative for this case; therefore, calculate the exact solution. Calculate U as the larger of the values from AISC Specification Section D3 and Table D3.1 Case 4. From AISC Specification Section D3, for open cross sections, U need not be less than the ratio of the gross area of the connected element(s) to the member gross area. U 

bf t f Ag

8.01 in. 0.515 in. 5.84 in.2

 0.706

Case 4, where l is the length of the connection and x  y :

3l 2

 x 1   3l  w  l  2    1.09 in.  3 16.0 in.  1   2 2   3 16.0 in.   8.01 in.   16.0 in.   0.860

U

2

2

Therefore, use U = 0.860. Calculate An using AISC Specification Section B4.3. Because there are no reductions due to bolt holes or notches: An  Ag  5.84 in.2

Calculate Ae using AISC Specification Section D3. Ae  AnU



 5.84 in.2

(Spec. Eq. D3-1)

  0.860

 5.02 in.2

Calculate Pn. Pn  Fu Ae



  65 ksi  5.02 in.2

(Spec. Eq. D2-2)



 326 kips

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From AISC Specification Section D2, the available tensile rupture strength is: LRFD t  0.75 

 t  2.00 

 t Pn  0.75  326 kips 



 245 kips  240 kips o.k.

ASD

Pn 326 kips  t 2.00  163 kips  160 kips o.k.



Alternately, the available tensile rupture strengths can be determined by modifying the tabulated values. The available tensile rupture strengths published in the tension member selection tables are based on the assumption that Ae = 0.75Ag. The actual available strengths can be determined by adjusting the values from AISC Manual Table 5-3 as follows: LRFD  Ae  t Pn   214 kips     0.75 Ag   5.02 in.2   214 kips    0.75 5.84 in.2   245 kips  240 kips o.k.



 Ae Pn  142 kips   t  0.75 Ag



   

ASD   

 5.02 in.2  142 kips    0.75 5.84 in.2   163 kips  160 kips o.k.



Recommended Slenderness Limit L  30.0 ft 12 in./ft   rx 1.57 in.  229  300 from AISC Specification Section D1 o.k.

Note: The L/rx limit is a recommendation, not a requirement. See Chapter J for illustrations of connection limit state checks.

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

   

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EXAMPLE D.4

RECTANGULAR HSS TENSION MEMBER

Given:

Verify the tensile strength of an ASTM A500 Grade C HSS64a with a length of 30 ft. The member is carrying a dead load of 40 kips and a live load of 110 kips in tension. As shown in Figure D.4-1, the end connection is a fillet welded 2-in.-thick single concentric gusset plate with a weld length of 16 in. Assume that the gusset plate and weld are satisfactory.

Fig. D.4-1. Connection geometry for Example D.4. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C, rectangular HSS Fy = 50 ksi Fu = 62 ksi From AISC Manual Table 1-11, the geometric properties are as follows: HSS64a Ag = 6.18 in.2 ry = 1.55 in. t = 0.349 in.

From Chapter 2 of ASCE/SEI 7, the required tensile strength is: LRFD Pu  1.2  40 kips   1.6 110 kips 

ASD Pa  40 kips  110 kips  150 kips

 224 kips Tensile Yielding

Check tensile yielding limit state using AISC Manual Table 5-4. LRFD t Pn  278 kips  224 kips

o.k. 

ASD Pn  185 kips  150 kips o.k. t

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Tensile Rupture Check tensile rupture limit state using AISC Manual Table 5-4. ASD

LRFD t Pn  216 kips  224 kips

Pn  144 kips  150 kips n.g. t

n.g. 

The tabulated available rupture strengths may be conservative in this case; therefore, calculate the exact solution. Calculate U from AISC Specification Section D3 and Table D3.1 Case 6. x 

B 2  2 BH 4B  H 

 4.00 in.2  2  4.00 in. 6.00 in. 4  4.00 in.  6.00 in.

 1.60 in. x l 1.60 in.  1 16.0 in.  0.900

U  1

Allowing for a z-in. gap in fit-up between the HSS and the gusset plate: An  Ag  2  t p  z in. t  6.18 in.2  2 2 in.  z in. 0.349 in.  5.79 in.2

Calculate Ae using AISC Specification Section D3. Ae  AnU



2

 5.79 in.

(Spec. Eq. D3-1)

  0.900

 5.21 in.2

Calculate Pn. Pn  Fu Ae



  62 ksi  5.21 in.2

(Spec. Eq. D2-2)



 323 kips

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From AISC Specification Section D2, the available tensile rupture strength is: LRFD t  0.75 

 t  2.00 

 t Pn  0.75  323 kips 



ASD

Pn 323 kips  t 2.00  162 kips  150 kips o.k.

 242 kips  224 kips o.k.

The HSS available tensile strength is governed by the tensile rupture limit state. Recommended Slenderness Limit

L  30.0 ft 12 in./ft   r 1.55 in.  232  300 from AISC Specification Section D1 o.k. Note: The L/r limit is a recommendation, not a requirement. See Chapter J for illustrations of connection limit state checks.

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EXAMPLE D.5

ROUND HSS TENSION MEMBER

Given: Verify the tensile strength of an ASTM A500 Grade C HSS6.0000.500 with a length of 30 ft. The member carries a dead load of 40 kips and a live load of 120 kips in tension. As shown in Figure D.5-1, the end connection is a fillet welded 2-in.-thick single concentric gusset plate with a weld length of 16 in. Assume that the gusset plate and weld are satisfactory.

Fig. D.5-1. Connection geometry for Example D.5. Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C, round HSS Fy = 46 ksi Fu = 62 ksi From AISC Manual Table 1-13, the geometric properties are as follows: HSS6.0000.500 Ag = 8.09 in.2 r = 1.96 in. t = 0.465 in.

From Chapter 2 of ASCE/SEI 7, the required tensile strength is: LRFD Pu  1.2  40 kips   1.6 120 kips 

ASD Pa  40 kips  120 kips  160 kips

 240 kips Tensile Yielding

Check tensile yielding limit state using AISC Manual Table 5-6. LRFD t Pn  335 kips  240 kips

o.k. 

ASD Pn  223 kips  160 kips o.k. t

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Tensile Rupture Check tensile rupture limit state using AISC Manual Table 5-6. LRFD t Pn  282 kips  240 kips

ASD Pn  188 kips  160 kips t

o.k. 

o.k.

Check that Ae Ag  0.75 as assumed in table. Determine U from AISC Specification Table D3.1 Case 5. l  16.0 in. D  6.00 in. l 16.0 in.  D 6.00 in.  2.67  1.3, therefore U  1.0 Allowing for a z-in. gap in fit-up between the HSS and the gusset plate, An  Ag  2  t p  z in. t  8.09 in.2  2 2 in.  z in. 0.465 in.  7.57 in.2

Calculate Ae using AISC Specification Section D3. Ae  AnU



(Spec. Eq. D3-1)



 7.57 in.2 1.0   7.57 in.2

Ae 7.57 in.2  Ag 8.09 in.2  0.936  0.75

o.k.

Because AISC Manual Table 5-6 provides an overly conservative estimate of the available tensile rupture strength for this example, calculate Pn using AISC Specification Section D2. Pn  Fu Ae



  62 ksi  7.57 in.

2

(Spec. Eq. D2-2)



 469 kips

From AISC Specification Section D2, the available tensile rupture strength is:

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LRFD t  0.75 

 t  2.00 

 t Pn  0.75  469 kips 



 352 kips  240 kips o.k.

ASD

Pn 469 kips  t 2.00  235 kips  160 kips o.k.



The HSS available strength is governed by the tensile yielding limit state. Recommended Slenderness Limit L  30.0 ft 12 in./ft   r 1.96 in.  184  300 from AISC Specification Section D1 o.k. Note: The L/r limit is a recommendation, not a requirement. See Chapter J for illustrations of connection limit state checks.

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EXAMPLE D.6

DOUBLE-ANGLE TENSION MEMBER

Given: An ASTM A36 2L442 (a-in. separation) has one line of eight w-in.-diameter bolts in standard holes and is 25 ft in length as shown in Figure D.6-1. The double angle is carrying a dead load of 40 kips and a live load of 120 kips in tension. Verify the member tensile strength. Assume that the gusset plate and bolts are satisfactory.

Fig. D.6-1. Connection geometry for Example D.6.

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-7 and 1-15, the geometric properties are as follows: L442

x = 1.18 in. 2L442 (s = a in.)

Ag = 7.50 in.2 ry = 1.83 in. rx = 1.21 in.

From Chapter 2 of ASCE/SEI 7, the required tensile strength is: LRFD Pu  1.2  40 kips   1.6 120 kips   240 kips

ASD Pa  40 kips  120 kips  160 kips

Tensile Yielding Check tensile yielding limit state using AISC Manual Table 5-8. LRFD t Pn  243 kips  240 kips o.k.

ASD Pn  162 kips  160 kips o.k. t

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Tensile Rupture Determine the available tensile rupture strength using AISC Specification Section D2. Calculate U as the larger of the values from AISC Specification Section D3, Table D3.1 Case 2 and Case 8. From AISC Specification Section D3, for open cross sections, U need not be less than the ratio of the gross area of the connected element(s) to the member gross area. Half of the member is connected, therefore, the minimum U value is:

U  0.500 From Case 2, where l is the length of connection: x l 1.18 in.  1 21.0 in.  0.944

U  1

From Case 8, with four or more fasteners per line in the direction of loading:

U  0.80 Therefore, use U = 0.944. Calculate An using AISC Specification Section B4.3. An  Ag  2  d h  z in. t  7.50 in.2  2 m in.  z in.2 in.  6.63 in.2

Calculate Ae using AISC Specification Section D3. Ae  AnU



 6.63 in.2

(Spec. Eq. D3-1)

  0.944 

 6.26 in.2

Calculate Pn. Pn  Fu Ae



  58 ksi  6.26 in.

2

(Spec. Eq. D2-2)



 363 kips

From AISC Specification Section D2, the available tensile rupture strength is:

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LRFD t  0.75 

 t  2.00 

 t Pn  0.75  363 kips 



ASD

Pn 363 kips  t 2.00  182 kips

 272 kips

Note that AISC Manual Table 5-8 could also be conservatively used since Ae  0.75Ag. The double-angle available tensile strength is governed by the tensile yielding limit state. LRFD 243 kips  240 kips o.k.

ASD 162 kips  160 kips o.k.

Recommended Slenderness Limit L  25.0 ft 12 in./ft   1.21 in. rx  248  300 from AISC Specification Section D1

o.k.

Note: From AISC Specification Section D4, the longitudinal spacing of connectors between components of built-up members should preferably limit the slenderness ratio in any component between the connectors to a maximum of 300. See Chapter J for illustrations of connection limit state checks.

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EXAMPLE D.7

PIN-CONNECTED TENSION MEMBER

Given: An ASTM A36 pin-connected tension member with the dimensions shown in Figure D.7-1 carries a dead load of 4 kips and a live load of 12 kips in tension. The diameter of the pin is 1 in., in a Q-in. oversized hole. Assume that the pin itself is adequate. Verify the member tensile strength.

Fig. D.7-1. Connection geometry for Example D.7.

Solution: From AISC Manual Table 2-5, the material properties are as follows: Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi The geometric properties of the plate are as follows: a b c d

 2.25 in.  1.61 in.  2.50 in.  1.00 in.

d h  1.03 in. t  2 in. w  4.25 in.

The requirements given in AISC Specification Sections D5.2(a) and D5.2(b) are satisfied by the given geometry. Requirements given in AISC Specification Sections D5.2(c) and D5.2(d) are checked as follows:

be  2t  0.63  b  2 2 in.  0.63  1.61 in.  1.63 in.  1.61 in. Therefore, use be = 1.61 in.

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a  1.33be 2.25 in.  1.33 1.61 in. 2.25 in.  2.14 in.

o.k.

w  2be  d 4.25 in.  2 1.61 in.  1.00 in. 4.25in.  4.22 in.

o.k.

ca 2.50 in.  2.25 in.

o.k.

From Chapter 2 of ASCE/SEI 7, the required tensile strength is: LRFD Pu  1.2  4 kips   1.6 12 kips 

ASD Pa  4 kips  12 kips

 16.0 kips

 24.0 kips

From AISC Specification Section D5.1, the available tensile strength is the lower value determined according to the limit states of tensile rupture, shear rupture, bearing and yielding. Tensile Rupture Calculate the available tensile rupture strength on the effective net area. Pn  Fu  2tbe 

(Spec. Eq. D5-1)

  58 ksi  2 2 in.1.61 in.  93.4 kips

From AISC Specification Section D5.1, the available tensile rupture strength is: LRFD t  0.75   t Pn  0.75  93.4 kips 

ASD

 t  2.00 

 Pn 93.4 kips  t 2.00  46.7 kips

 70.1 kips Shear Rupture

From AISC Specification Section D5.1, the area on the shear failure path is: d  Asf  2t  a   2    1.00 in.    2 2 in.  2.25 in.     2    2.75 in.2

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Pn  0.6 Fu Asf

(Spec. Eq. D5-2)



 0.6  58 ksi  2.75 in.2



 95.7 kips From AISC Specification Section D5.1, the available shear rupture strength is: LRFD

ASD

sf  0.75 

sf  2.00 

 sf Pn  0.75  95.7 kips 

 Pn 95.7 kips   sf 2.00

 71.8 kips

 47.9 kips

Bearing Determine the available bearing strength using AISC Specification Section J7.

Apb  td   2 in.1.00 in.  0.500 in.2 Rn  1.8Fy Apb

(Spec. Eq. J7-1)



 1.8  36 ksi  0.500 in.

2



 32.4 kips From AISC Specification Section J7, the available bearing strength is: LRFD

  0.75 

 Pn  0.75  32.4 kips   24.3 kips

  2.00   Pn 32.4 kips   2.00  16.2 kips

ASD

Tensile Yielding Determine the available tensile yielding strength using AISC Specification Section D2(a). Ag  wt   4.25 in.2 in.  2.13 in.2 Pn  Fy Ag

(Spec. Eq. D2-1)



  36 ksi  2.13 in.

2



 76.7 kips

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From AISC Specification Section D2, the available tensile yielding strength is: LRFD t  0.90 

 t  1.67 

 t Pn  0.90  76.7 kips 



ASD

Pn 76.7 kips  t 1.67  45.9 kips

 69.0 kips

The available tensile strength is governed by the bearing strength limit state. LRFD Pn  24.3 kips  24.0 kips o.k.

ASD Pn  16.2 kips  16.0 kips o.k. 

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EXAMPLE D.8

EYEBAR TENSION MEMBER

Given: A s-in.-thick, ASTM A36 eyebar member as shown in Figure D.8, carries a dead load of 25 kips and a live load of 15 kips in tension. The pin diameter, d, is 3 in. Verify the member tensile strength.

Fig. D.8-1. Connection geometry for Example D.8.

Solution: From AISC Manual Table 2-5, the material properties are as follows: Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi The geometric properties of the eyebar are as follows: R

 8.00 in.

b

 2.23 in.

d

 3.00 in.

dh

 3.03 in.

d head  7.50 in. t

 s in.

w

 3.00 in.

Check the dimensional requirement using AISC Specification Section D6.1.

w  8t 3.00 in.  8  s in. 3.00 in.  5.00 in. o.k. Check the dimensional requirements using AISC Specification Section D6.2. t  2 in.

s in.  2 in. o.k.

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7 w 8 7 3.00 in.   3.00 in. 8 3.00 in.  2.63 in. o.k. d

d h  d  Q in. 3.03 in.  3.00 in.  Q in. 3.03 in.  3.03 in.

o.k.

R  d head 8.00 in.  7.50 in. o.k. 2 3 wb w 3 4 2 3  3.00 in.  2.23 in.   3.00 in. 3 4 2.00 in.  2.23 in.  2.25 in. o.k.

From Chapter 2 of ASCE/SEI 7, the required tensile strength is: LRFD Pu  1.2  25 kips   1.6 15 kips 

ASD Pa  25 kips  15 kips  40.0 kips

 54.0 kips Tensile Yielding

Determine the available tensile yielding strength using AISC Specification Section D2 at the eyebar body (at w). Ag  wt   3.00 in. s in.  1.88 in.2 Pn  Fy Ag

(Spec. Eq. D2-1)



  36 ksi  1.88 in.2



 67.7 kips The available tensile yielding strength is: LRFD t  0.90 

 t  1.67 

 t Pn  0.90  67.7 kips 



 60.9 kips  54.0 kips o.k.

ASD

Pn 67.7 kips  t 1.67  40.5 kips  40.0 kips

o.k.

The eyebar tension member available strength is governed by the tensile yielding limit state.

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Note: The eyebar detailing limitations ensure that the tensile yielding limit state at the eyebar body will control the strength of the eyebar itself. The pin should also be checked for shear yielding, and, if the material strength is less than that of the eyebar, the bearing limit state should also be checked.

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EXAMPLE D.9

PLATE WITH STAGGERED BOLTS

Given: Compute An and Ae for a 14-in.-wide and 2-in.-thick plate subject to tensile loading with staggered holes as shown in Figure D.9-1.

Fig. D.9-1. Connection geometry for Example D.9.

Solution: Calculate the net hole diameter using AISC Specification Section B4.3b. d net  d h  z in.  m in.  z in.  0.875 in.

Compute the net width for all possible paths across the plate. Because of symmetry, many of the net widths are identical and need not be calculated. w  14.0 in.  d net  

s2 from AISC Specification Section B4.3b. 4g

Line A-B-E-F: w  14.0 in.  2  0.875 in.  12.3 in.

Line A-B-C-D-E-F: w  14.0 in.  4  0.875 in. 

 2.50 in.2  2.50 in.2  4  3.00 in. 4  3.00 in.

 11.5 in.

Line A-B-C-D-G:

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 2.50in.2 w  14.0 in. 3  0.875in.  4  3.00 in.  11.9 in.

Line A-B-D-E-F: w  14.0 in.  3  0.875 in. 

 2.50 in.2  2.50 in.2  4  7.00 in. 4  3.00 in.

 12.1 in.

Line A-B-C-D-E-F controls the width, w, therefore: An  wt  11.5 in.2 in.  5.75 in.2

Calculate U. From AISC Specification Table D3.1 Case 1, because tension load is transmitted to all elements by the fasteners, U = 1.0

Ae  AnU



(Spec. Eq. D3-1)



 5.75 in.2 1.0   5.75 in.2

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Chapter E Design of Members for Compression This chapter covers the design of compression members, the most common of which are columns. The AISC Manual includes design tables for the following compression member types in their most commonly available grades:      

W-shapes and HP-shapes Rectangular, square and round HSS Pipes WT-shapes Double angles Single angles

LRFD and ASD information is presented side-by-side for quick selection, design or verification. All of the tables account for the reduced strength of sections with slender elements. The design and selection method for both LRFD and ASD is similar to that of previous editions of the AISC Specification, and will provide similar designs. In this AISC Specification, LRFD and ASD will provide identical designs when the live load is approximately three times the dead load. The design of built-up shapes with slender elements can be tedious and time consuming, and it is recommended that standard rolled shapes be used whenever possible. E1. GENERAL PROVISIONS The design compressive strength, cPn, and the allowable compressive strength, Pn/c, are determined as follows: Pn = nominal compressive strength is the lowest value obtained based on the applicable limit states of flexural buckling, torsional buckling, and flexural-torsional buckling, kips c = 0.90 (LRFD)

c = 1.67 (ASD)

Because the critical stress, Fcr, is used extensively in calculations for compression members, it has been tabulated in AISC Manual Table 4-14 for all of the common steel yield strengths. E2. EFFECTIVE LENGTH In the AISC Specification, there is no limit on slenderness, Lc/r. Per the User Note in AISC Specification Section E2, it is recommended that Lc/r not exceed 200, as a practical limit based on professional judgment and construction economics. Although there is no restriction on the unbraced length of columns, the tables of the AISC Manual are stopped at common or practical lengths for ordinary usage. For example, a double L334, with a a-in. separation has an ry of 1.38 in. At a Lc/r of 200, this strut would be 23 ft long. This is thought to be a reasonable limit based on fabrication and handling requirements. Throughout the AISC Manual, shapes that contain slender elements for compression when supplied in their most common material grade are footnoted with the letter “c.” For example, see a W1422c.

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E3. FLEXURAL BUCKLING OF MEMBERS WITHOUT SLENDER ELEMENTS Nonslender-element compression members, including nonslender built-up I-shaped columns and nonslender HSS columns, are governed by these provisions. The general design curve for critical stress versus Lc/r is shown in Figure E-1. The term Lc is used throughout this chapter to describe the length between points that are braced against lateral and/or rotational displacement. E4. TORSIONAL AND FLEXURAL-TORSIONAL BUCKLING OF SINGLE ANGLES AND MEMBERS WITHOUT SLENDER ELEMENTS This section is most commonly applicable to double angles and WT sections, which are singly symmetric shapes subject to torsional and flexural-torsional buckling. The available strengths in axial compression of these shapes are tabulated in AISC Manual Part 4 and examples on the use of these tables have been included in this chapter for the shapes. E5. SINGLE-ANGLE COMPRESSION MEMBERS The available strength of single-angle compression members is tabulated in AISC Manual Part 4. E6. BUILT-UP MEMBERS The available strengths in axial compression for built-up double angles with intermediate connectors are tabulated in AISC Manual Part 4. There are no tables for other built-up shapes in the AISC Manual, due to the number of possible geometries. E7. MEMBERS WITH SLENDER ELEMENTS The design of these members is similar to members without slender elements except that a reduced effective area is used in lieu of the gross cross-sectional area. The tables of AISC Manual Part 4 incorporate the appropriate reductions in available strength to account for slender elements. Design examples have been included in this Chapter for built-up I-shaped members with slender webs and slender flanges. Examples have also been included for a double angle, WT and an HSS with slender elements.

Fig. E-1. Standard column curve.

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Table E-1 Limiting Values of Lc /r and Fe Fy, ksi

Limiting Lc / r

Fe, ksi

36

134

15.9

50

113

22.4

65

99.5

28.9

70

95.9

31.1

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EXAMPLE E.1A W-SHAPE COLUMN DESIGN WITH PINNED ENDS Given: Select a W-shape column to carry the loading as shown in Figure E.1A. The column is pinned top and bottom in both axes. Limit the column size to a nominal 14-in. shape. A column is selected for both ASTM A992 and ASTM A913 Grade 65 material.

Fig. E.1A. Column loading and bracing. Solution: Note that ASTM A913 Grade 70 might also be used in this design. The requirement for higher preheat when welding and the need to use 90-ksi filler metals for complete-joint-penetration (CJP) welds to other 70-ksi pieces offset the advantage of the lighter column and should be considered in the selection of which grade to use. From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi ASTM A913 Grade 65 Fy = 65 ksi From ASCE/SEI 7, Chapter 2, the required compressive strength is: LRFD

Pu  1.2 140 kips   1.6  420 kips   840 kips

ASD

Pa  140 kips  420 kips  560 kips

Column Selection—ASTM A992 From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, Kx = Ky = 1.0. The effective length is:

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Lc  K x Lx  K y Ly  1.0  30 ft   30.0 ft

Because the unbraced length is the same in both the x-x and y-y directions and rx exceeds ry for all W-shapes, y-y axis bucking will govern. Enter AISC Manual Table 4-1a with an effective length, Lc, of 30 ft, and proceed across the table until reaching the least weight shape with an available strength that equals or exceeds the required strength. Select a W14132. From AISC Manual Table 4-1a, the available strength for a y-y axis effective length of 30 ft is: LRFD c Pn  893 kips  840 kips

ASD

Pn  594 kips  560 kips o.k. c

o.k. 

Column Selection–ASTM A913 Grade 65 Enter AISC Manual Table 4-1b with an effective length, Lc, of 30 ft, and proceed across the table until reaching the least weight shape with an available strength that equals or exceeds the required strength. Select a W14120. From AISC Manual Table 4-1b, the available strength for a y-y axis effective length of 30 ft is: LRFD c Pn  856 kips  840 kips

o.k. 

ASD Pn  569 kips  560 kips o.k. c

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EXAMPLE E.1B W-SHAPE COLUMN DESIGN WITH INTERMEDIATE BRACING Given:

Verify a W1490 is adequate to carry the loading as shown in Figure E.1B. The column is pinned top and bottom in both axes and braced at the midpoint about the y-y axis and torsionally. The column is verified for both ASTM A992 and ASTM A913 Grade 65 material.

Fig. E.1B. Column loading and bracing. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi ASTM A913 Grade 65 Fy = 65 ksi From ASCE/SEI 7, Chapter 2, the required compressive strength is: LRFD Pu  1.2 140 kips   1.6  420 kips   840 kips

ASD

Pa  140 kips  420 kips  560 kips

From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, Kx = Ky = 1.0. The effective length about the y-y axis is: Lcy  K y Ly  1.0 15 ft   15.0 ft

The values tabulated in AISC Manual Tables 4-1a, 4-1b and 4-1c are provided for buckling in the y-y direction. To determine the buckling strength in the x-x axis, an equivalent effective length for the y-y axis is determined using the rx/ry ratio provided at the the bottom of these tables. For a W1490, rx/ry = 1.66, and the equivalent y-y axis effective length for x-x axis buckling is computed as:

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Lcx  K x Lx  1.0  30 ft   30.0 ft Lcy eq 

Lcx rx ry

(Manual Eq. 4-1)

30.0 ft 1.66  18.1 ft 

Because 18.1 ft > 15.0 ft, the available compressive strength is governed by the x-x axis flexural buckling limit state. Available Compressive Strength—ASTM A992 The available strength of a W1490 is determined using AISC Manual Table 4-1a, conservatively using an unbraced length of Lc = 19.0 ft. LRFD c Pn  903 kips  840 kips

ASD

Pn  601 kips  560 kips o.k. c

o.k. 

Available Compressive Strength—ASTM 913 Grade 65 The available strength of a W1490 is determined using AISC Manual Table 4-1b, conservatively using an unbraced length of Lc = 19.0 ft. ASD

LRFD c Pn  1, 080 kips  840 kips

o.k. 

Pn  719 kips  560 kips o.k. c

The available strengths of the columns described in Examples E.1A and E.1B are easily selected directly from the AISC Manual Tables. The available strengths can also be determined as shown in the following Examples E.1C and E.1D.

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EXAMPLE E.1C W-SHAPE AVAILABLE STRENGTH CALCULATION Given:

Calculate the available strength of the column sizes selected in Example E.1A with unbraced lengths of 30 ft in both axes. The material properties and loads are as given in Example E.1A. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi ASTM A913 Grade 65 Fy = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W14120 Ag = 35.3 in.2 rx = 6.24 in. ry = 3.74 in. W14132 Ag = 38.8 in.2 rx = 6.28 in. ry = 3.76 in.

Column Compressive Strength—ASTM A992 Slenderness Check From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, Kx = Ky = 1.0. The effective length about the y-y axis is: Lcy  K y Ly  1.0  30 ft   30.0 ft

Because the unbraced length for the W14132 column is the same for both axes, the y-y axis will govern. Lcy  30.0 ft 12 in./ft   3.76 in. ry  95.7

Critical Stress For Fy = 50 ksi, the available critical stresses, cFcr and Fcr/c for Lc/r = 95.7 are interpolated from AISC Manual Table 4-14 as follows. The available critical stress can also be determined as shown in Example E.1D.

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LRFD

ASD

Fcr  15.4 ksi c

c Fcr  23.0 ksi

From AISC Specification Equation E3-1, the available compressive strength of the W14132 column is: c Pn   c Fcr  Ag

ASD

LRFD

  23.0 ksi   38.8 in.

2

Pn  Fcr   Ag  c   c 



 892 kips  840 kips

 15.4 ksi   38.8 in.2 

o.k.

 598 kips  560 kips

o.k.

Column Compressive Strength—ASTM A913 Grade 65 Slenderness Check From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, Kx = Ky = 1.0. The effective length about the y-y axis is: Lcy  K y Ly  1.0  30 ft   30.0 ft

Because the unbraced length for the W14120 column is the same for both axes, the y-y axis will govern. Lcy  30.0 ft 12 in./ft   3.74 in. ry  96.3

Critical Stress For Fy = 65 ksi, the available critical stresses, cFcr and Fcr/c for Lc/r = 96.3 are interpolated from AISC Manual Table 4-14 as follows. The available critical stress can also be determined as shown in Example E.1D. LRFD

ASD

Fcr  16.1 ksi c

c Fcr  24.3 ksi

From AISC Specification Equation E3-1, the available compressive strength of the W14120 column is: c Pn   c Fcr  Ag

ASD

LRFD Pn  Fcr   c   c

  24.3 ksi   35.3 in.2   858 kips  840 kips

o.k.

  Ag 

 16.1 ksi   35.3 in.2   568 kips  560 kips

Note that the calculated values are approximately equal to the tabulated values.

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EXAMPLE E.1D W-SHAPE AVAILABLE STRENGTH CALCULATION Given:

Calculate the available strength of a W1490 with a x-x axis unbraced length of 30 ft and y-y axis and torsional unbraced lengths of 15 ft. The material properties and loads are as given in Example E.1A. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi ASTM A913 Grade 65 Fy = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W1490

Ag = 26.5 in.2 rx = 6.14 in. ry = 3.70 in. bf = 10.2 2t f

h = 25.9 tw Slenderness Check From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, Kx = Ky = 1.0. Lcx  K x Lx  1.0  30 ft   30.0 ft Lcx  30.0 ft 12 in./ft   6.14 in. rx  58.6 governs Lcy  K y Ly  1.0 15 ft   15.0 ft Lcy 15.0 ft 12 in./ft   3.70 in. ry  48.6

Column Compressive Strength—ASTM A992 Width-to-Thickness Ratio

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The width-to-thickness ratio of the flanges of the W1490 is:

bf  10.2 2t f From AISC Specification Table B4.1a, Case 1, the limiting width-to-thickness ratio of the flanges is: 0.56

E 29, 000 ksi  0.56 50 ksi Fy  13.5  10.2; therefore, the flanges are nonslender

The width-to-thickness ratio of the web of the W1490 is:

h  25.9 tw From AISC Specification Table B4.1a, Case 5, the limiting width-to-thickness ratio of the web is: 1.49

E 29, 000 ksi  1.49 50 ksi Fy  35.9  25.9; therefore, the web is nonslender

Because the web and flanges are nonslender, the limit state of local buckling does not apply. Critical Stresses The available critical stresses may be interpolated from AISC Manual Table 4-14 or calculated directly as follows. Calculate the elastic critical buckling stress, Fe, according to AISC Specification Section E3. As noted in AISC Specification Commentary Section E4, torsional buckling of symmetric shapes is a failure mode usually not considered in the design of hot-rolled columns. This failure mode generally does not govern unless the section is manufactured from relatively thin plates or a torsional unbraced length significantly larger than the y-y axis flexural unbraced length is present. Fe 



2 E  Lc     r 

(Spec. Eq. E3-4)

2

2  29, 000 ksi 

 58.6 2

 83.3 ksi

Calculate the flexural buckling stress, Fcr. 4.71

E 29, 000 ksi  4.71 50 ksi Fy  113

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Because

Lc  58.6  113, r

Fy   Fcr  0.658 Fe  

  Fy  

(Spec. Eq. E3-2)

50 ksi     0.65883.3 ksi   50 ksi      38.9 ksi

Nominal Compressive Strength Pn  Fcr Ag

(Spec. Eq. E3-1)



  38.9 ksi  26.5 in.2



 1, 030 kips

From AISC Specification Section E1, the available compressive strength is: LRFD

ASD c  1.67   Pn 1, 030 kips   c 1.67  617 kips  560 kips o.k.

c  0.90 

 c Pn  0.90 1, 030 kips   927 kips  840 kips o.k.



Column Compressive Strength—ASTM A913 Grade 65 Width-to-Thickness Ratio The width-to-thickness ratio of the flanges of the W1490 is:

bf  10.2 2t f From AISC Specification Table B4.1a, Case 1, the limiting width-to-thickness ratio of the flanges is: 0.56

E 29, 000 ksi  0.56 65 ksi Fy

 11.8  10.2; therefore, the flanges are nonslender

The width-to-thickness ratio of the web of the W1490 is:

h  25.9 tw From AISC Specification Table B4.1a, Case 5, the limiting width-to-thickness ratio of the web is:

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1.49

E 29, 000 ksi  1.49 65 ksi Fy

 31.5  25.9; therefore, the web is nonslender

Because the web and flanges are nonslender, the limit state of local buckling does not apply. Critical Stress Fe  83.3 ksi (calculated previously)

Calculate the flexural buckling stress, Fcr. E 29, 000 ksi  4.71 65 ksi Fy

4.71

 99.5

Because

Lc  58.6  99.5, r

Fy  Fcr   0.658 Fe  

  Fy  

(Spec. Eq. E3-2)

65 ksi     0.65883.3 ksi   65 ksi       46.9 ksi

Nominal Compressive Strength Pn  Fcr Ag

(Spec. Eq. E3-1)



  46.9 ksi  26.5 in.

2



 1, 240 kips

From AISC Specification Section E1, the available compressive strength is: LRFD c  0.90 

 c Pn  0.90 1, 240 kips   1,120 kips  840 kips o.k.



ASD c  1.67   Pn 1, 240 kips   c 1.67  743 kips  560 kips o.k.

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EXAMPLE E.2

BUILT-UP COLUMN WITH A SLENDER WEB

Given:

Verify that a built-up, ASTM A572 Grade 50 column with PL1 in. 8 in. flanges and a PL4 in. 15 in. web, as shown in Figure E2-1, is sufficient to carry a dead load of 70 kips and live load of 210 kips in axial compression. The column’s unbraced length is 15 ft and the ends are pinned in both axes.

Fig. E.2-1. Column geometry for Example E.2. Solution:

From AISC Manual Table 2-5, the material properties are as follows: Built-Up Column ASTM A572 Grade 50 Fy = 50 ksi Fu = 65 ksi The geometric properties are as follows: Built-Up Column d = 17.0 in. bf = 8.00 in. tf = 1.00 in. h = 15.0 in. tw = 4 in. From ASCE/SEI 7, Chapter 2, the required compressive strength is: LRFD Pu  1.2  70 kips   1.6  210 kips 

ASD

Pa  70 kips  210 kips  280 kips

 420 kips Built-Up Section Properties (ignoring fillet welds)

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Ag  2b f t f  htw  2  8.00 in.1.00 in.  15.0 in.4 in.  19.8 in.2 Iy  

bh3 12

 1.00 in. 8.00 in.3  15.0 in.4 in.3   2 12 12    85.4 in.4

Iy

ry 

A 85.4 in.4



19.8 in.2  2.08 in. I x   Ad 2  

bh3 12

4 in.15.0 in.3   8.00 in.1.00 in.3  2 +2  2  8.00 in.2  8.00 in.  +   12 12  





 1,100 in.4

Elastic Flexural Buckling Stress From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, Ky = 1.0. Because the unbraced length is the same for both axes, the y-y axis will govern by inspection. With Lcy = KyLy = 1.0(15 ft) = 15.0 ft: Lcy ry



15.0 ft 12 in./ft  2.08 in.

 86.5 Fe 



2 E  Lcy     ry 

(from Spec. Eq. E3-4)

2

2  29, 000 ksi 

 86.5 2

 38.3 ksi

Elastic Critical Torsional Buckling Stress Note: Torsional buckling generally will not govern for doubly symmetric members if Lcy  Lcz ; however, the check is included here to illustrate the calculation.

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From the User Note in AISC Specification Section E4: Cw 

I y ho 2 4

85.4 in.  16.0 in. 4



2

4 6

 5, 470 in.

From AISC Design Guide 9, Equation 3.4: J 

bt 3 3

  8.00 in.1.00 in.3  15.0 in.4 in.3   2 3 3    5.41 in.4

 2 ECw  1 Fe   + GJ  2  Lcz  Ix  I y 6  2   1    29, 000 ksi  5, 470 in. 4  + 11, 200 ksi 5.41in.     2 4 4   1.0 15 ft 12 in./ft     1,100 in.  85.4 in.   91.9 ksi  38.3 ksi







(Spec. Eq. E4-2)



Therefore, the flexural buckling limit state controls. Use Fe = 38.3 ksi. Flexural Buckling Stress Fy 50 ksi  Fe 38.3 ksi  1.31

Fy  2.25, Fe

Because

Fy  Fcr   0.658 Fe  



  Fy  

(Spec. Eq. E3-2)



 0.6581.31  50 ksi   28.9 ksi

Slenderness Check for slender flanges using AISC Specification Table B4.1a.

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Calculate kc using AISC Specification Table B4.1a, note [a]. kc  

4 h tw 4

15.0 in. 4 in.  0.516, which is between 0.35 and 0.76.

For the flanges: b t 4.00 in.  1.00 in.  4.00



Determine the flange limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 2: kc E Fy

 r  0.64

0.516  29, 000 ksi 

 0.64

50 ksi

 11.1

Because    r , the flanges are not slender and there is no reduction in effective area due to local buckling of the flanges. Check for a slender web, and then determine the effective area for compression, Ae, using AISC Specification Section E7.1. h tw 15.0 in.  4 in.  60.0



Determine the slender web limit from AISC Specification Table B4.1a, Case 5:  r  1.49  1.49

E Fy 29, 000 ksi 50 ksi

 35.9

Because    r , the web is slender. Determine the slenderness limit from AISC Specification Section E7.1 for a fully effective element:

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r

Fy Fcr

50 ksi 28.9 ksi

 35.9  47.2

Fy , the effective width is determined from AISC Specification Equation E7-3. Determine the Fcr effective width imperfection adjustment factors from AISC Specification Table E7.1, Case (a):

Because    r

c1  0.18 c2  1.31 The elastic local buckling stress is: 2

   Fel   c2 r  Fy      35.9    1.31   60.0     30.7 ksi

(Spec. Eq. E7-5) 2

 50 ksi 

Determine the effective width of the web and the resulting effective area:  F  F he  h 1  c1 el  el Fcr  Fcr   30.7 ksi  30.7 ksi  15.0 in. 1  0.18  28.9 ksi  28.9 ksi   12.6 in.

(from Spec. Eq. E7-3)

Ae  Ag   h  he  tw  19.8 in.2  15.0 in.  12.6 in.4 in.  19.2 in.2

Available Compressive Strength Pn  Fcr Ae



  28.9 ksi  19.2 in.2

(Spec. Eq. E7-1)



 555 kips

From AISC Specification Section E1, the available compressive strength is: LRFD

ASD

c  0.90

c  1.67

c Pn  0.90  555 kips 

Pn 555 kips  c 1.67  332 kips  280 kips o.k.

 500 kips  420 kips o.k.

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EXAMPLE E.3

BUILT-UP COLUMN WITH SLENDER FLANGES

Given:

Determine if a built-up, ASTM A572 Grade 50 column with PLa in. 102 in. flanges and a PL4 in. 74 in. web, as shown in Figure E.3-1, has sufficient available strength to carry a dead load of 40 kips and a live load of 120 kips in axial compression. The column’s unbraced length is 15 ft and the ends are pinned in both axes.

Fig. E.3-1. Column geometry for Example E.3. Solution:

From AISC Manual Table 2-5, the material properties are as follows: Built-Up Column ASTM A572 Grade 50 Fy = 50 ksi Fu = 65 ksi The geometric properties are as follows: Built-Up Column d = 8.00 in. bf = 102 in. tf = a in. h = 74 in. tw = 4 in. From ASCE/SEI 7, Chapter 2, the required compressive strength is: LRFD Pu  1.2  40 kips   1.6 120 kips 

ASD

Pa  40 kips  120 kips  160 kips

 240 kips Built-Up Section Properties (ignoring fillet welds) Ag  2 102 in. a in.   74 in.4 in.  9.69 in.2

Because the unbraced length is the same for both axes, the weak axis will govern.

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Iy  

bh3 12

  a in.102 in.3   74 in.4 in.3   2 12 12    72.4 in.4

ry  

Iy Ag 72.4 in.4

9.69 in.2  2.73 in. I x   Ad 2  

bh3 12

4 in. 74 in.3  102 in. a in.3  2  +2  2 102 in. a in. 3.81 in. +   12 12    122 in.4

Web Slenderness Determine the limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 5:  r  1.49  1.49

E Fy 29, 000 ksi 50 ksi

 35.9

h tw 74 in.  4 in.  29.0



Because    r , the web is not slender. Note that the fillet welds are ignored in the calculation of h for built up sections. Flange Slenderness Calculate kc using AISC Specification Table B4.1a, note [a]:

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kc  

4 h tw 4

74 in. 4 in.  0.743, which is between 0.35 and 0.76

Determine the limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 2:  r  0.64  0.64

kc E Fy 0.743  29, 000 ksi  50 ksi

 13.3 b t 5.25 in.  a in.  14.0



Because    r , the flanges are slender. For compression members with slender elements, AISC Specification Section E7 applies. The nominal compressive strength, Pn, is determined based on the limit states of flexural, torsional and flexural-torsional buckling. Depending on the slenderness of the column, AISC Specification Equation E3-2 or E3-3 applies. Fe is used in both equations and is calculated as the lesser of AISC Specification Equations E3-4 and E4-2. From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0. Because the unbraced length is the same for both axes, the weak axis will govern. With Lcy = KyLy = 1.0(15 ft) = 15.0 ft: Lcy 15.0 ft 12 in./ft   2.73 in. ry  65.9

Elastic Critical Stress, Fe, for Flexural Buckling Fe 



2 E  Lcy     ry 

(from Spec. Eq. E3-4)

2

2  29, 000 ksi 

 65.9 2

 65.9 ksi

Elastic Critical Stress, Fe, for Torsional Buckling

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Note: This limit state is not likely to govern, but the check is included here for completeness. From the User Note in AISC Specification Section E4: Cw 

I y ho 2 4

 72.4 in.   7.63 in. 4



2

4

 1, 050 in.

6

From AISC Design Guide 9, Equation 3.4: J 

bt 3 3

2 102 in. a in. +  74 in.4 in. 3



3

3

 0.407 in.4

With Lcz = KzLz = 1.0(15 ft) = 15 ft:  2 ECw  1 Fe   + GJ  2  Lcz  Ix  Iy



(Spec. Eq. E4-2)



 2  29, 000 ksi 1, 050 in.6   + 11, 200 ksi 0.407 in.4 2 15 ft 12 in./ft     71.2 ksi  65.9 ksi





   122 in. 

4

   72.4 in.4  1

Therefore, use Fe = 65.9 ksi. Flexural Buckling Stress Fy 50 ksi  Fe 65.9 ksi  0.759

Fy  2.25 : Fe

Because

Fy   Fcr  0.658 Fe  



  Fy  

(Spec. Eq. E3-2)



 0.6580.759  50 ksi   36.4 ksi

Effective Area, Ae

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The effective area, Ae, is the summation of the effective areas of the cross section based on the reduced effective widths, be or he. Since the web is nonslender, there is no reduction in the effective area due to web local buckling and he = h. Determine the slender web limit from AISC Specification Section E7.1. r

Fy Fcr

50 ksi 36.4 ksi

 13.3  15.6

Because    r

Fy Fcr

for all elements,

be  b

(Spec. Eq. E7-2)

Therefore, Ae  Ag . Available Compressive Strength Pn  Fcr Ae



  36.4 ksi  9.69 in.2



(Spec. Eq. E7-1)

 353 kips

From AISC Specification Section E1, the available compressive strength is: LRFD

ASD

c = 0.90

c = 1.67

c Pn  0.90  353 kips 

Pn 353 kips  1.67 c  211 kips  160 kips o.k.

 318 kips  240 kips o.k.

Note: Built-up sections are generally more expensive than standard rolled shapes; therefore, a standard compact shape, such as a W835 might be a better choice even if the weight is somewhat higher. This selection could be taken directly from AISC Manual Table 4-1a.

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EXAMPLE E.4A W-SHAPE COMPRESSION MEMBER (MOMENT FRAME)

This example is primarily intended to illustrate the use of the alignment chart for sidesway uninhibited columns in conjunction with the effective length method. Given:

The member sizes shown for the moment frame illustrated here (sidesway uninhibited in the plane of the frame) have been determined to be adequate for lateral loads. The material for both the column and the girders is ASTM A992. The loads shown at each level are the accumulated dead loads and live loads at that story. The column is fixed at the base about the x-x axis of the column. Determine if the column is adequate to support the gravity loads shown. Assume the column is continuously supported in the transverse direction (the y-y axis of the column). Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W1850 Ix = 800 in.4 W2455 Ix = 1,350 in.4 W1482 Ag = 24.0 in.2 Ix = 881 in.4

Column B-C From ASCE/SEI 7, Chapter 2, the required compressive strength for the column between the roof and floor is: LRFD Pu  1.2  41.5 kips   1.6 125 kips   250 kips

ASD Pa  41.5 kips  125 kips  167 kips

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Effective Length Factor Using the effective length method, the effective length factor is determined using AISC Specification Commentary Appendix 7, Section 7.2. As discussed there, column inelasticity should be addressed by incorporating the stiffness reduction parameter, b. Determine Gtop and Gbottom accounting for column inelasticity by replacing EcolIcol with b(EcolIcol). Calculate the stiffness reduction parameter, τb, for the column B-C using AISC Manual Table 4-13. LRFD

ASD Pa 167 kips = Ag 24.0 in.2  6.96 ksi

Pu 250 kips  Ag 24.0 in.2  10.4 ksi

b  1.00

b  1.00

Therefore, no reduction in stiffness for inelastic buckling will be required. Determine Gtop and Gbottom.   ( EI / L)col  Gtop  b     ( EI / L) g 

(from Spec. Comm. Eq. C-A-7-3)





   29, 000 ksi  881 in.4           14.0 ft     1.00    4     29, 000 ksi  800 in.   2   35.0 ft       1.38





  ( EI / L)col  Gbottom  b     ( EI / L) g 

(from Spec. Comm. Eq. C-A-7-3)

   29, 000 ksi   881 in.4     2   14.0 ft      1.00   4    29, 000 ksi  1,350 in.     2  35.0 ft      1.63

From the alignment chart, AISC Specification Commentary Figure C-A-7.2, K is slightly less than 1.5; therefore use K = 1.5. Because the column available strength tables are based on the Lc about the y-y axis, the equivalent effective column length of the upper segment for use in the table is: Lcx   KL  x

 1.5 14 ft   21.0 ft

From AISC Manual Table 4-1a, for a W1482:

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rx  2.44 ry Lcx  rx     ry  21.0 ft  2.44  8.61 ft

Lc 

Take the available strength of the W1482 from AISC Manual Table 4-1a. At Lc = 9 ft, the available strength in axial compression is: LRFD c Pn  940 kips > 250 kips o.k.

ASD

Pn  626 kips > 167 kips o.k. c

Column A-B From Chapter 2 of ASCE/SEI 7, the required compressive strength for the column between the floor and the foundation is: LRFD Pu  1.2 100 kips   1.6  300 kips   600 kips

ASD

Pa  100 kips  300 kips  400 kips

Effective Length Factor Determine the stiffness reduction parameter, τb, for column A-B using AISC Manual Table 4-13. LRFD

ASD

Pu 600 kips  Ag 24.0 in.2  25.0 ksi

Pa 400 kips = Ag 24.0 in.2  16.7 ksi

b  1.00

b  0.994

Use b = 0.994.

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   EI / L   col  Gtop  b     EI / L  g 

(from Spec. Comm. Eq. C-A-7-3)





   29, 000 ksi  881 in.4    2       14.0 ft     0.994    4     29, 000 ksi  1,350 in.   2  35.0 ft       1.62





Gbottom  1.0  fixed  , from AISC Specification Commentary Appendix 7, Section 7.2 From the alignment chart, AISC Specification Commentary Figure C-A-7.2, K is approximately 1.4. Because the column available strength tables are based on Lc about the y-y axis, the effective column length of the lower segment for use in the table is:

Lcx   KL  x

 1.4 14 ft   19.6 ft

Lc 

Lcx

 rx     ry  19.6 ft  2.44  8.03 ft

Take the available strength of the W1482 from AISC Manual Table 4-1a. At Lc = 9 ft, (conservative) the available strength in axial compression is: LRFD c Pn  940 kips > 600 kips o.k.

ASD

Pn  626 kips > 400 kips o.k. c

A more accurate strength could be determined by interpolation from AISC Manual Table 4-1a.

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EXAMPLE E.4B W-SHAPE COMPRESSION MEMBER (MOMENT FRAME) Given:

Using the effective length method, determine the available strength of the column shown subject to the same gravity loads shown in Example E.4A with the column pinned at the base about the x-x axis. All other assumptions remain the same.

Solution:

As determined in Example E.4A, for the column segment B-C between the roof and the floor, the column strength is adequate. As determined in Example E.4A, for the column segment A-B between the floor and the foundation,

Gtop  1.62 At the base, Gbottom  10 (pinned) from AISC Specification Commentary Appendix 7, Section 7.2

Note: this is the only change in the analysis. From the alignment chart, AISC Specification Commentary Figure C-A-7.2, K is approximately equal to 2.0. Because the column available strength tables are based on the effective length, Lc, about the y-y axis, the effective column length of the segment A-B for use in the table is: Lcx   KL  x

 2.0 14 ft   28.0 ft

From AISC Manual Table 4-1a, for a W1482:

rx  2.44 ry

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Lc 

Lcx

 rx     ry  28.0 ft  2.44  11.5 ft

Interpolate the available strength of the W14×82 from AISC Manual Table 4-1a. LRFD c Pn  861 kips > 600 kips o.k.

ASD Pn  573 kips > 400 kips o.k. c

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EXAMPLE E.5

DOUBLE-ANGLE COMPRESSION MEMBER WITHOUT SLENDER ELEMENTS

Given:

Verify the strength of a 2L432a LLBB (w-in. separation) strut, ASTM A36, with a length of 8 ft and pinned ends carrying an axial dead load of 20 kips and live load of 60 kips. Also, calculate the required number of pretensioned bolted or welded intermediate connectors required. The solution will be provided using: (1) AISC Manual Tables (2) Calculations using AISC Specification provisions Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-7 and 1-15, the geometric properties are as follows: L432a rz = 0.719 in. 2L432a LLBB

rx = 1.25 in. ry = 1.55 in. for a-in. separation ry = 1.69 in. for w-in. separation From ASCE/SEI 7, Chapter 2, the required compressive strength is: LRFD Pu  1.2  20 kips   1.6  60 kips   120 kips

ASD

Pa  20 kips  60 kips  80.0 kips

(1) AISC Manual Table Solution From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0. Therefore, Lcx = Lcy = KL = 1.0(8 ft) = 8.00 ft. The available strength in axial compression is taken from the upper (X-X Axis) portion of AISC Manual Table 4-9: LRFD c Pn  127 kips > 120 kips o.k.

ASD Pn  84.7 kips > 80.0 kips o.k. c

For buckling about the y-y axis, the values are tabulated for a separation of a in. To adjust to a spacing of w in., Lcy is multiplied by the ratio of the ry for a a-in. separation to the ry for a w-in. separation, where Lcy = KyLy = 1.0(8 ft) = 8.00 ft . Thus:  1.55 in.  Lcy   8.00 ft     1.69 in.   7.34 ft

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The calculation of the equivalent Lcy in the preceding text is a simplified approximation of AISC Specification Section E6.1. To ensure a conservative adjustment for a w-in. separation, take Lcy = 8 ft. The available strength in axial compression is taken from the lower (Y-Y Axis) portion of AISC Manual Table 4-9 as: LRFD c Pn  132 kips > 120 kips

ASD

Pn  87.9 kips > 80.0 kips o.k. c

o.k.

Therefore, x-x axis flexural buckling governs. Intermediate Connectors From AISC Manual Table 4-9, at least two welded or pretensioned bolted intermediate connectors are required. This can be verified as follows: a  distance between connectors 

8.00 ft 12 in./ft 

3 spaces  32.0 in. From AISC Specification Section E6.2, the effective slenderness ratio of the individual components of the built-up member based upon the distance between intermediate connectors, a, must not exceed three-fourths of the governing slenderness ratio of the built-up member. Therefore,

a 3  Lc     . ri 4  r max

Solving for a gives: L  3ri  c   r max a 4 Lcx  8.00 ft 12 in./ft   1.25 in. rx  76.8 controls

Lcy  8.00 ft 12 in./ft   ry 1.69 in.  56.8 L  3rz  c   r  max a 4 3  0.719in. 76.8   4  41.4 in.

Therefore, two welded or pretensioned bolted connectors are adequate since 32.0 in. < 41.4 in.

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Note that one connector would not be adequate as 48.0 in. > 41.4 in. Available strength can also be determined by hand calculations, as demonstrated in the following. (2) Calculations Using AISC Specification Provisions From AISC Manual Tables 1-7 and 1-15, the geometric properties are as follows: L432a J = 0.132 in.4 2L432a LLBB (w in. separation)

Ag = 5.36 in.2 ry = 1.69 in. ro  2.33 in. H = 0.813

Slenderness Check b t 4.00 in.  a in.  10.7



Determine the limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 3:  r  0.45



 0.45

E Fy 29, 000 ksi 36 ksi

 12.8    r ; therefore, there are no slender elements.

For double-angle compression members without slender elements, AISC Specification Sections E3, E4 and E6 apply. The nominal compressive strength, Pn, is determined based on the limit states of flexural, torsional and flexuraltorsional buckling. Flexural Buckling about the x-x Axis Lcx  8.00 ft 12 in./ft   1.25 in. rx  76.8

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Fex 



2 E  Lcx   r   x 

(Spec. Eq. E4-5)

2

2  29, 000 ksi 

 76.82

 48.5 ksi

Flexural Buckling about the y-y Axis Lcy  8.00 ft 12 in./ft   ry 1.69 in.  56.8

Using AISC Specification Section E6, compute the modified Lc/r for built up members with pretensioned bolted or welded connectors. Assume two connectors are required. a

8.00 ft 12 in./ft  3

 32.0 in. ri  rz (single angle)  0.719 in. a 32.0 in.  ri 0.719 in.  44.5  40

Therefore: 2

 Ki a   Lc   Lc          r m  r o  ri 

2

(Spec. Eq. E6-2b)

where Ki = 0.50 for angles back-to-back

 Lc      r m

 0.50  32.0 in.    0.719 in. 

 56.82  

2

 61.0 Fey 



2 E  Lcy     ry 

(Spec. Eq. E4-6)

2

2  29, 000 ksi 

 61.0 2

 76.9 ksi

Torsional and Flexural-Torsional Buckling

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For nonslender double-angle compression members, AISC Specification Equation E4-3 applies. Per the User Note for AISC Specification Section E4, the term with Cw is omitted when computing Fez and xo is taken as zero. The flexural buckling term about the y-y axis, Fey, was computed in the preceding section.  2 ECw  1 Fez    GJ  2 2  Lcz  Ag ro



  0  11, 200 ksi  0.132 in.4 

(Spec. Eq. E4-7)

  2 angles 

1

5.36 in.   2.33 in. 2

2

 102 ksi  4 Fey Fez H    1  1  2   Fey  Fez     76.9 ksi  102 ksi   4  76.9 ksi 102 ksi  0.813     1  1  2  0.813     76.9 ksi  102 ksi 2   60.5 ksi

 Fey  Fez Fe    2H

(Spec. Eq. E4-3)

Critical Buckling Stress The critical buckling stress for the member could be controlled by flexural buckling about either the x-x axis or y-y axis, Fex or Fey, respectively. Note that AISC Specification Equations E4-5 and E4-6 reflect the same buckling modes as calculated in AISC Specification Equation E3-4. Or, the critical buckling stress for the member could be controlled by torsional or flexural-torsional buckling calculated per AISC Specification Equation E4-3. In this example, Fe calculated in accordance with AISC Specification Equation E4-5 (or Equation E3-4) is less than that calculated in accordance with AISC Specification Equation E4-3 or E4-6, and controls. Therefore: Fe  48.5 ksi Fy 36 ksi  Fe 48.5 ksi  0.742

Per the AISC Specification User Note for Section E3, the two inequalities for calculating limits of applicability of Sections E3(a) and E3(b) provide the same result for flexural buckling only. When the elastic buckling stress, Fe, is controlled by torsional or flexural-torsional buckling, the Lc/r limits would not be applicable unless an equivalent Lc/r ratio is first calculated by substituting the governing Fe into AISC Specification Equation E3-4 and solving for Lc/r. The Fy/Fe limits may be used regardless of which buckling mode governs.

Fy  2.25 : Fe

Because

Fy  Fcr   0.658 Fe  



  Fy  

(Spec. Eq. E3-2)



 0.6580.742  36 ksi   26.4 ksi

Available Compressive Strength

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Pn  Fcr Ag

(Spec. Eq. E3-1, Eq. E4-1)



  26.4 ksi  5.36 in.

2



 142 kips

From AISC Specification Section E1, the available compressive strength is: ASD

LRFD c = 0.90

c = 1.67

c Pn  0.90 142 kips 

Pn 142 kips  c 1.67  85.0 kips  80.0 kips

 128 kips  120 kips o.k.

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EXAMPLE E.6

DOUBLE-ANGLE COMPRESSION MEMBER WITH SLENDER ELEMENTS

Given:

Determine if a 2L534 LLBB (w-in. separation) strut, ASTM A36, with a length of 8 ft and pinned ends has sufficient available strength to support a dead load of 10 kips and live load of 30 kips in axial compression. Also, calculate the required number of pretensioned bolted or welded intermediate connectors. The solution will be provided using: (1) AISC Manual Tables (2) Calculations using AISC Specification provisions Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-7 and 1-15, the geometric properties are as follows: L534 rz = 0.652 in. 2L534 LLBB

rx = 1.62 in. ry = 1.19 in. for a-in. separation ry = 1.33 in. for w-in. separation From ASCE/SEI 7, Chapter 2, the required compressive strength is: LRFD Pu  1.2 10 kips   1.6  30 kips   60.0 kips

ASD

Pa  10 kips  30 kips  40.0 kips

(1) AISC Manual Table Solution

From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0. Therefore, Lcx = Lcy = KL = 1.0(8 ft) = 8.00 ft. The available strength in axial compression is taken from the upper (X-X Axis) portion of AISC Manual Table 4-9: ASD

LRFD c Pnx  91.2 kips > 60.0 kips o.k.

Pnx  60.7 kips > 40.0 kips o.k. c

For buckling about the y-y axis, the tabulated values are based on a separation of a in. To adjust for a spacing of w in., Lcy is multiplied by the ratio of ry for a a-in. separation to ry for a w-in. separation.  1.19 in.  Lcy   8.00 ft     1.33 in.   7.16 ft

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This calculation of the equivalent Lcy does not completely take into account the effect of AISC Specification Section E6.1 and is slightly unconservative. From the lower portion of AISC Manual Table 4-9, interpolate for a value at Lcy = 7.16 ft. The available strength in compression is: LRFD

c Pny  68.3 kips > 60.0 kips o.k.

ASD Pny  45.4 kips > 40.0 kips o.k. c

These strengths are approximate due to the linear interpolation from the table and the approximate value of the equivalent Lcy noted in the preceding text. These can be compared to the more accurate values calculated in detail as follows. Intermediate Connectors From AISC Manual Table 4-9, it is determined that at least two welded or pretensioned bolted intermediate connectors are required. This can be confirmed by calculation, as follows: a  distance between connectors 

8.00 ft 12 in./ft 

3 spaces  32.0 in. From AISC Specification Section E6.2, the effective slenderness ratio of the individual components of the built-up member based upon the distance between intermediate connectors, a, must not exceed three-fourths of the governing slenderness ratio of the built-up member. Therefore,

a 3  Lc     . ri 4  r  max

Solving for a gives: L  3ri  c   r max a 4 ri  rz  0.652 in. Lcx  8.00 ft 12 in./ft   1.62 in. rx  59.3

Lcy  8.00 ft 12 in./ft   ry 1.33 in.  72.2

controls

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L  3rz  c   r  max a 4 3  0.652 in. 72.2   4  35.3 in.

Therefore, two welded or pretensioned bolted connectors are adequate since 32.0 in. < 35.3 in. Available strength can also be determined by hand calculations, as determined in the following. (2) Calculations Using AISC Specification Provisions From AISC Manual Tables 1-7 and 1-15, the geometric properties are as follows. L534 J = 0.0438 in.4 rz = 0.652 in. 2L534 LLBB

Ag = 3.88 in.2 rx = 1.62 in. ry = 1.33 in. for w-in. separation ro  2.59 in. H = 0.657 Slenderness Check For the 5-in. leg: b t 5.00 in.  4 in.  20.0



For the 3-in. leg: b t 3.00 in.  4 in.  12.0



Calculate the limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 3:  r  0.45  0.45

E Fy 29, 000 ksi 36 ksi

 12.8

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For the longer leg,    r , and therefore it is classified as a slender element. For the shorter leg,    r , and therefore it is classified as a nonslender element. For a double-angle compression member with slender elements, AISC Specification Section E7 applies. The nominal compressive strength, Pn, is determined based on the limit states of flexural, torsional and flexural-torsional buckling. Ae will be determined by AISC Specification Section E7.1. Elastic Buckling Stress about the x-x Axis With Lcx = KxLx = 1.0(8 ft) = 8.00 ft: Lcx  8.00 ft 12 in./ft   1.62 in. rx  59.3

Fex 



2 E  Lcx   r   x 

(Spec. Eq. 3-4 or E4-5)

2

2  29, 000 ksi 

 59.32

 81.4

Elastic Buckling Stress about the y-y Axis With Lcy = KyLy = 1.0(8 ft) = 8.00 ft: Lcy  8.00 ft 12 in./ft   ry 1.33 in.  72.2

Using AISC Specification Section E6, compute the modified Lcy/ry for built-up members with pretensioned bolted or welded connectors. Assuming two connectors are required: a

8.00 ft 12 in./ft  3

 32.0 in. ri  rz (single angle)  0.652 in. a 32.0 in.  ri 0.652 in.  49.1  40

Therefore: 2

 Ki a   Lc   Lc          r m  r o  ri 

2

(Spec. Eq. E6-2b)

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where Ki = 0.50 for angles back-to-back  Lc      r m

 0.50  32.0 in.    0.652 in. 

 72.2 2  

2

 76.3

Fey 



2 E  Lcy     ry 

(Spec. Eq. E3-4 or E4-6)

2

2  29, 000 ksi 

 76.32

 49.2 ksi Torsional and Flexural-Torsional Elastic Buckling Stress Per the User Note in AISC Specification Section E4, the term with Cw is omitted when computing Fez, and xo is taken as zero. The flexural buckling term about the y-y axis, Fey, was computed in the preceding section.  2 ECw  1  GJ  Fez   2 2  Lcz  Ag ro



  0  11, 200 ksi  0.0438 in.4 

(Spec. Eq. E4-7)

  2 angles 

1

 3.88 in.   2.59 in. 2

2

 37.7 ksi

4 Fey Fez H     1  1  2    Fey  Fez    49.2 ksi  37.7 ksi   4  49.2 ksi  37.7 ksi  0.657      1  1  2 2  0.657       49.2 ksi  37.7 ksi   26.8 ksi controls

 Fey  Fez Fe    2H

(Spec. Eq. E4-3)

Critical Buckling Stress The critical buckling stress for the member could be controlled by flexural buckling about either the x-x axis or y-y axis, Fex or Fey, respectively. Note that AISC Specification Equations E4-5 and E4-6 reflect the same buckling modes as calculated in AISC Specification Equation E3-4. Or, the critical buckling stress for the member could be controlled by torsional or flexural-torsional buckling calculated per AISC Specification Equation E4-3. In this example, Fe calculated in accordance with AISC Specification Equation E4-3 is less than that calculated in accordance with AISC Specification Equation E4-5 or E4-6, and controls. Therefore: Fe  26.8 ksi Fy 36 ksi  Fe 26.8 ksi  1.34

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Per the AISC Specification User Note for Section E3, the two inequalities for calculating limits of applicability of Sections E3(a) and E3(b) provide the same result for flexural buckling only. When the elastic buckling stress, Fe, is controlled by torsional or flexural-torsional buckling, the Lc/r limits would not be applicable unless an equivalent Lc/r ratio is first calculated by substituting the governing Fe into AISC Specification Equation E3-4 and solving for Lc/r. The Fy/Fe limits may be used regardless of which buckling mode governs.

Fy  2.25 : Fe

Because

Fy   Fcr  0.658 Fe  



  Fy  

(Spec. Eq. E3-2)



 0.6581.34  36 ksi   20.5 ksi

Effective Area Determine the limits of applicability for local buckling in accordance with AISC Specification Section E7.1. The shorter leg was shown previously to be nonslender and therefore no reduction in effective area due to local buckling of the shorter leg is required. The longer leg was shown previously to be slender and therefore the limits of AISC Specification Section E7.1 need to be evaluated.   20.0

r

Fy 36 ksi  12.8 20.5 ksi Fcr  17.0 Fy , determine the effective width imperfection adjustment factors per AISC Specification Table Fcr

Because    r E7.1, Case (c).

c1  0.22 c2  1.49 Determine the elastic local buckling stress from AISC Specification Section E7.1. 2

   Fel   c2 r  Fy   

(Spec. Eq. E7-5) 2

  12.8    1.49     36 ksi   20.0     32.7 ksi Determine the effective width of the angle leg and the resulting effective area.

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 F  F be  b 1  c1 el  el Fcr  Fcr   32.7 ksi  32.7 ksi   5.00 in. 1  0.22  20.5 ksi  20.5 ksi   4.56 in.

(Spec. Eq. E7-3)

Ae  Ag  t   b  be 





 3.88 in.2  4 in. 5.00 in.  4.56 in. 2 angles  2

 3.66 in.

Available Compressive Strength Pn  Fcr Ae



  20.5 ksi  3.66 in.

2

(Spec. Eq. E7-1)



 75.0 kips

From AISC Specification Section E1, the available compressive strength is: LRFD

ASD

c = 0.90

c = 1.67

c Pn  0.90  75.0 kips 

Pn 75.0 kips  c 1.67  44.9 kips  40.0 kips

 67.5 kips  60.0 kips o.k.



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EXAMPLE E.7

WT COMPRESSION MEMBER WITHOUT SLENDER ELEMENTS

Given:

Select an ASTM A992 nonslender WT-shape compression member with a length of 20 ft to support a dead load of 20 kips and live load of 60 kips in axial compression. The ends are pinned. The solution will be provided using: (1) AISC Manual Tables (2) Calculations using AISC Specification provisions Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From ASCE/SEI 7, Chapter 2, the required compressive strength is: LRFD Pu  1.2  20 kips   1.6  60 kips 

ASD

Pa  20 kips  60 kips  80.0 kips

 120 kips (1) AISC Manual Table Solution

From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0. Therefore, Lcx = Lcy = KL = 1.0(20 ft) = 20.0 ft. Select the lightest nonslender member from AISC Manual Table 4-7 with sufficient available strength about both the x-x axis (upper portion of the table) and the y-y axis (lower portion of the table) to support the required strength. Try a WT734. The available strength in compression is: LRFD c Pnx  128 kips  120 kips

o.k. controls

ASD Pnx  85.5 kips  80.0 kips o.k. controls c

Pny  147 kips  80.0 kips o.k. c

c Pny  222 kips  120 kips o.k.

Available strength can also be determined by hand calculations, as demonstrated in the following. (2) Calculation Using AISC Specification Provisions

From AISC Manual Table 1-8, the geometric properties are as follows. WT734

Ag = 10.0 in.2 rx = 1.81 in. ry = 2.46 in. J = 1.50 in.4

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y = 1.29 in. Ix = 32.6 in.4 Iy = 60.7 in.4 d = 7.02 in. tw = 0.415 in. bf = 10.0 in. tf = 0.720 in.

Stem Slenderness Check d tw 7.02in.  0.415in.



 16.9 Determine the stem limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 4:  r  0.75  0.75

E Fy 29, 000 ksi 50 ksi

 18.1    r ; therefore, the stem is not slender

Flange Slenderness Check 

bf 2t f

10.0 in. 2(0.720 in.)  6.94

=

Determine the flange limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 1:  r  0.56  0.56

E Fy 29,000 ksi 50 ksi

 13.5

   r ; therefore, the flange is not slender

There are no slender elements. For compression members without slender elements, AISC Specification Sections E3 and E4 apply. The nominal compressive strength, Pn, is determined based on the limit states of flexural, torsional and flexural-torsional buckling.

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Elastic Flexural Buckling Stress about the x-x Axis Lcx  20.0 ft 12 in./ft   1.81 in. rx  133

Fex 



2 E  Lcx   r   x 

(Spec. Eq. E3-4 or E4-5)

2

2  29, 000 ksi 

1332

 16.2 ksi

controls

Elastic Flexural Buckling Stress about the y-y Axis Lcy  20.0 ft 12 in./ft   ry 2.46 in.  97.6

Fey 



2 E  Lcy     ry 

(Spec. Eq. E3-4 or E4-6)

2

2  29, 000 ksi 

 97.6 2

 30.0 ksi Torsional and Flexural-Torsional Elastic Buckling Stress Because the WT734 section does not have any slender elements, AISC Specification Section E4 will be applicable for torsional and flexural-torsional buckling. Fe will be calculated using AISC Specification Equation E4-3. Per the User Note for AISC Specification Section E4, the term with Cw is omitted when computing Fez, and xo is taken as zero. The flexural buckling term about the y-y axis, Fey, was computed in the preceding section. xo  0

yo  y 

tf 2

 1.29 in. 

0.720 in. 2

 0.930 in.

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ro 2  xo 2  yo 2 

Ix  I y Ag

(Spec. Eq. E4-9)

32.6 in.4  60.7 in.4

 0   0.930 in.  2

10.0 in.2

2

 10.2 in.

 2 ECw  1  GJ  Fez   2 2  Lcz  Ag ro



(Spec. Eq. E4-7)



1   0  11, 200 ksi  1.50 in.4    10.0 in.2 10.2 in.2







 165 ksi H  1  1

xo 2  yo 2

(Spec. Eq. E4-8)

ro 2 0   0.930 in.

2

10.2 in.2

 0.915  4 Fey Fez H    1  1  2   Fey  Fez     30.0 ksi  165 ksi   4  30.0 ksi 165 ksi  0.915      1  1  2  0.915      30.0 ksi  165 ksi 2   29.5 ksi

 Fey  Fez Fe    2H

(Spec. Eq. E4-3)

Critical Buckling Stress The critical buckling stress for the member could be controlled by flexural buckling about either the x-x axis or y-y axis, Fex or Fey, respectively. Note that AISC Specification Equations E4-5 and E4-6 reflect the same buckling modes as calculated in AISC Specification Equation E3-4. Or, the critical buckling stress for the member could be controlled by torsional or flexural-torsional buckling calculated per AISC Specification Equation E4-3. In this example, Fe calculated in accordance with AISC Specification Equation E4-5 is less than that calculated in accordance with AISC Specification Equation E4-3 or E4-6 and controls. Therefore: Fe  16.2 ksi Fy 50 ksi  Fe 16.2 ksi  3.09

Per the AISC Specification User Note for Section E3, the two inequalities for calculating limits of applicability of Sections E3(a) and E3(b) provide the same result for flexural buckling only. When the elastic buckling stress, Fe, is controlled by torsional or flexural-torsional buckling, the Lc/r limits would not be applicable unless an equivalent Lc/r ratio is first calculated by substituting the governing Fe into AISC Specification Equation E3-4 and solving for Lc/r. The Fy/Fe limits may be used regardless of which buckling mode governs.

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Because

Fy  2.25 : Fe

Fcr  0.877 Fe

(Spec. Eq. E3-3)

 0.877 16.2 ksi   14.2 ksi

Available Compressive Strength Pn  Fcr Ag

(Spec. Eq. E3-1)



 14.2 ksi  10.0 in.

2



 142 kips

From AISC Specification Section E1, the available compressive strength is: LRFD

ASD

c  0.90

c  1.67

c Pn  0.90 142 kips 

Pn 142 kips  c 1.67  85.0 kips  80.0 kips

 128 kips  120 kips o.k.

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EXAMPLE E.8

WT COMPRESSION MEMBER WITH SLENDER ELEMENTS

Given: Select an ASTM A992 WT-shape compression member with a length of 20 ft to support a dead load of 6 kips and live load of 18 kips in axial compression. The ends are pinned. The solution will be provided using: (1) AISC Manual Tables (2) Calculations using AISC Specification provisions

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From ASCE/SEI 7, Chapter 2 , the required compressive strength is: ASD

LRFD Pu  1.2  6 kips   1.6 18 kips 

Pa  6 kips  18 kips  24.0 kips

 36.0 kips (1) AISC Manual Table Solution

From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0. Therefore, Lcx = Lcy = KL = 1.0(20 ft) = 20.0 ft. Select the lightest member from AISC Manual Table 4-7 with sufficient available strength about the both the x-x axis (upper portion of the table) and the y-y axis (lower portion of the table) to support the required strength. Try a WT715. The available strength in axial compression from AISC Manual Table 4-7 is: ASD Pnx  49.4 kips  24.0 kips o.k. c

LRFD c Pnx  74.3 kips  36.0 kips

o.k.

c Pny  36.6 kips  36.0 kips

o.k. controls

Pny  24.4 kips  24.0 kips o.k. controls c

Available strength can also be determined by hand calculations, as demonstrated in the following. (2) Calculation Using AISC Specification Provisions

From AISC Manual Table 1-8, the geometric properties are as follows: WT715

Ag = 4.42 in.2 rx = 2.07 in. ry = 1.49 in.

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J = 0.190 in.4 y = 1.58 in. Ix = 19.0 in.4 Iy = 9.79 in.4 d = 6.92 in. tw = 0.270 in. bf = 6.73 in. tf = 0.385 in. Stem Slenderness Check

d tw 6.92 in. = 0.270 in.  25.6



Determine stem limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 4:  r  0.75  0.75

E Fy 29, 000 ksi 50 ksi

 18.1    r ; therefore, the stem is slender

Flange Slenderness Check  

bf 2t f 6.73 in. 2  0.385 in.

 8.74

Determine flange limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 1:  r  0.56  0.56

E Fy 29, 000 ksi 50 ksi

 13.5    r ; therefore, the flange is not slender

Because this WT715 has a slender web, AISC Specification Section E7 is applicable. The nominal compressive strength, Pn, is determined based on the limit states of flexural, torsional and flexural-torsional buckling. Elastic Flexural Buckling Stress about the x-x Axis

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Lcx  20.0 ft 12 in./ft   rx 2.07 in.  116

Fex 



2 E  Lcx   r   x 

(Spec. Eq. E3-4 or E4-5)

2

2  29, 000 ksi 

116 2

 21.3

Elastic Flexural Buckling Stress about the y-y Axis Lcy ry



 20.0 ft 12 in./ft  1.49 in.

 161

Fey 



2 E  Lcy     ry 

(Spec. Eq. E3-4 or E4-6)

2

2  29, 000 ksi 

1612

 11.0 ksi Torsional and Flexural-Torsional Elastic Buckling Stress Fe will be calculated using AISC Specification Equation E4-3. Per the User Note for AISC Specification Section E4, the term with Cw is omitted when computing Fez, and xo is taken as zero. The flexural buckling term about the y-y axis, Fey, was computed in the preceding section. xo  0 yo  y 

tf 2

 1.58 in. 

0.385 in. 2

 1.39 in. ro 2  xo 2  yo 2 

Ix  I y Ag

 0  1.39 in.  2

(Spec. Eq. E4-9)

19.0 in.4  9.79 in.4 4.42 in.2

 8.45 in.2

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 2 ECw  1 Fez    GJ  2 2  Lcz  Ag ro



(Spec. Eq. E4-7)



1   0  11, 200 ksi  0.190 in.4    4.42 in.2 8.45 in.2







 57.0 ksi H  1  1

xo 2  yo 2

(Spec. Eq. E4-8)

ro 2 0  1.39 in.

2

8.45 in.2

 0.771  Fey  Fez Fe    2H

 4 Fey Fez H   1  1  2   Fey  Fez  

   

(Spec. Eq. E4-3)

11.0 ksi  57.0 ksi   4 11.0 ksi  57.0 ksi  0.771       1 1  2  0.771 11.0 ksi  57.0 ksi 2      10.5 ksi controls

Critical Buckling Stress The critical buckling stress for the member could be controlled by flexural buckling about either the x-x axis or y-y axis, Fex or Fey, respectively. Note that AISC Specification Equations E4-5 and E4-6 reflect the same buckling modes as calculated in AISC Specification Equation E3-4. Or, the critical buckling stress for the member could be controlled by torsional or flexural-torsional buckling calculated per AISC Specification Equation E4-3. In this example, Fe calculated in accordance with AISC Specification Equation E4-3 is less than that calculated in accordance with AISC Specification Equation E4-5 or E4-6 and controls. Therefore: Fe  10.5 ksi Fy Fe

50 ksi 10.5 ksi  4.76 

Per the AISC Specification User Note for Section E3, the two inequalities for calculating limits of applicability of Sections E3(a) and E3(b) provide the same result for flexural buckling only. When the elastic buckling stress, Fe, is controlled by torsional or flexural-torsional buckling, the Lc /r limits would not be applicable unless an equivalent Lc/r ratio is first calculated by substituting the governing Fe into AISC Specification Equation E3-4 and solving for Lc/r. The Fy/Fe limits may be used regardless of which buckling mode governs. Because

Fy  2.25 : Fe

Fcr  0.877 Fe

(Spec. Eq. E3-3)

 0.877 10.5 ksi   9.21 ksi

Effective Area

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Because this section was found to have a slender element, the limits of AISC Specification Section E7.1 must be evaluated to determine if there is a reduction in effective area due to local buckling. Since the flange was found to not be slender, no reduction in effective area due to local buckling in the flange is required. Only a reduction in effective area due to local buckling in the stem may be required.

  25.6 r

Fy Fcr

50 ksi 9.21 ksi

 18.1  42.2

Because    r

Fy Fcr

,

be  b

(Spec. Eq. E7-2)

There is no reduction in effective area due to local buckling of the stem at the critical stress level and Ae = Ag. Available Compressive Strength Pn  Fcr Ae



  9.21 ksi  4.42 in.

2

(Spec. Eq. E7-1)



 40.7 kips

From AISC Specification Section E1, the available compressive strength is: ASD

LRFD c = 0.90

c = 1.67

c Pn  0.90  40.7 kips 

Pn 40.7 kips  c 1.67  24.4 kips  24.0 kips

 36.6 kips  36.0 kips o.k.

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EXAMPLE E.9 RECTANGULAR HSS COMPRESSION MEMBER WITHOUT SLENDER ELEMENTS Given: Select an ASTM A500 Grade C rectangular HSS compression member, with a length of 20 ft, to support a dead load of 85 kips and live load of 255 kips in axial compression. The base is fixed and the top is pinned. The solution will be provided using: (1) AISC Manual Tables (2) Calculations using AISC Specification provisions

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C, rectangular HSS Fy = 50 ksi Fu = 62 ksi From ASCE/SEI 7, Chapter 2, the required compressive strength is: ASD

LRFD Pu  1.2 85 kips   1.6  255 kips 

Pa  85 kips  255 kips  340 kips

 510 kips (1) AISC Manual Table Solution

From AISC Specification Commentary Table C-A-7.1, for a fixed-pinned condition, Kx = Ky = 0.80. Lc  K x Lx  K y Ly  0.80  20 ft   16.0 ft

Enter AISC Manual Table 4-3 for rectangular sections. Try a HSS1210a. From AISC Manual Table 4-3, the available strength in axial compression is: ASD Pn  370 kips  340 kips o.k. c

LRFD c Pn  556 kips  510 kips

o.k.

Available strength can also be determined by hand calculations, as demonstrated in the following. (2) Calculation Using AISC Specification Provisions

From AISC Manual Table 1-11, the geometric properties are as follows:

HSS1210a

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Ag = 14.6 in.2 t = 0.349 in. rx = 4.61 in. ry = 4.01 in. b/t = 25.7 h/t = 31.4 Slenderness Check Determine the wall limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 6: E Fy

 r  1.40

29, 000 ksi 50 ksi

 1.40  33.7

For the narrow side:   b t  25.7

For the wide side:   h t  31.4

   r ; therefore, the section does not contain slender elements.

Elastic Buckling Stress Because ry < rx and Lcx = Lcy, ry will govern the available strength. Determine the applicable equation: Lcy 16.0 ft 12 in./ft   ry 4.01 in.  47.9

4.71

E 29, 000 ksi  4.71 Fy 50 ksi  113  47.9

Therefore, use AISC Specification Equation E3-2. Fe 



2 E  Lc     r 

(Spec. Eq. E3-4)

2

2 (29, 000 ksi)

 47.9 2

 125 ksi

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Critical Buckling Stress Fy  Fcr   0.658 Fe  

  Fy  

(Spec. Eq. E3-2)

50 ksi     0.658125 ksi   50 ksi     42.3 ksi

Available Compressive Strength Pn  Fcr Ag

(Spec. Eq. E3-1)



  42.3 ksi  14.6 in.2



 618 kips

From AISC Specification Section E1, the available compressive strength is: LRFD c = 0.90 c Pn  0.90  618 kips   556 kips  510 kips o.k.



ASD c = 1.67  Pn 618 kips   c 1.67  370 kips  340 kips o.k.

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EXAMPLE E.10 RECTANGULAR HSS COMPRESSION MEMBER WITH SLENDER ELEMENTS Given:

Using the AISC Specification provisions, calculate the available strength of a HSS128x compression member with an effective length of Lc = 24 ft with respect to both axes. Use ASTM A500 Grade C. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C, rectangular HSS Fy = 50 ksi Fu = 62 ksi From AISC Manual Table 1-11 the geometric properties of an HSS128x are as follows: A  6.76 in.2 t  0.174 in. rx  4.56 in. ry  3.35 in. b  43.0 t h  66.0 t Slenderness Check Calculate the limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 6 for walls of rectangular HSS.

 r  1.40  1.40

E Fy 29, 000 ksi 50 ksi

 33.7 Determine the width-to-thickness ratios of the HSS walls. For the narrow side: b t  43.0   r  33.7



For the wide side: h t  66.0   r  33.7



All walls of the HSS128x are slender elements and the provisions of AISC Specification Section E7 apply.

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Critical Stress, Fcr From AISC Specification Section E7, the critical stress, Fcr, is calculated using the gross section properties and following the provisions of AISC Specification Section E3. The effective slenderness ratio about the y-axis will control. From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0. Therefore, Lcy = KyLy = 1.0(24 ft) = 24.0 ft. Lcy  Lc     ry  r max 

 24.0 ft 12 in./ft  3.35 in.

 86.0 4.71

E 29, 000 ksi  4.71 Fy 50 ksi  113  86.0

Therefore, use AISC Specification Equation E3-2.

Fe 



2 E  Lc     r 

(Spec. Eq. E3-4)

2

2  29, 000 ksi 

 86.0 2

 38.7 ksi Fy   Fcr  0.658 Fe  

  Fy  

(Spec. Eq. E3-2)

 50 ksi        0.658 38.7 ksi    50 ksi       29.1 ksi

Effective Area, Ae Compute the effective wall widths, he and be, in accordance with AISC Specification Section E7.1. Compare  for each wall with the following limit to determine if a local buckling reduction applies.

r

Fy 50 ksi  33.7 29.1 ksi Fcr  44.2

For the narrow walls: b t  43.0  44.2



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Therefore, the narrow wall width does not need to be reduced (be = b) per AISC Specification Equation E7-2. For the wide walls: h t  66.0  44.2



h Therefore, use AISC Specification Equation E7-3, with h    t   66.0  0.174 in.  11.5 in. t

The effective width imperfection adjustment factors, c1 and c2, are selected from AISC Specification Table E7.1, Case (b):

c1  0.20 c2  1.38 2

   Fel   c2 r  Fy      33.7    1.38    66.0     24.8 ksi

(Spec. Eq. E7-5) 2

 50 ksi 

 F  F he  h 1  c1 el  el Fcr  Fcr 

(Spec. Eq. E7-3)

 24.8 ksi  24.8 ksi  11.5 in. 1  0.20  29.1 ksi  29.1 ksi   8.66 in.

The effective area, Ae, is determined using the effective width he = 8.66 in. and the design wall thickness t = 0.174 in. As shown in Figure E.10-1, h – he is the width of the wall segments that must be reduced from the gross area, A, to compute the effective area, Ae. Note that a similar deduction would be required for the narrow walls if be  b.

Fig. E.10-1. HSS Effective Area. Ae  A  2  h  he  t  6.76 in.2  2 11.5 in.  8.66 in. 0.174 in.  5.77 in.2

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Available Compressive Strength The effective area is used to compute nominal compressive strength:

Pn  Fcr Ae



  29.1 ksi  5.77 in.

2

(Spec. Eq. E7-1)



 168 kips From AISC Specification Section E1, the available compressive strength is: LRFD

ASD

c  0.90

 c  1.67

c Pn  0.90 168 kips 

Pn 168 kips  c 1.67  101 kips

 151 kips Discussion The width-to-thickness criterion,  r  1.40

E for HSS in Table B4.1a is based on the assumption that the element Fy

will be stressed to Fy. If the critical flexural buckling stress is less than Fy, which it always is for compression members of reasonable length, wall local buckling may or may not occur before member flexural buckling occurs. For the case where the flexural buckling stress is low enough, wall local buckling will not occur. This is the case addressed in AISC Specification Section E7.1(a). For members where the flexural buckling stress is high enough, wall local buckling will occur. This is the case addressed in AISC Specification Section E7.1(b). The HSS128x in this example is slender according to Table B4.1a. For effective length Lc = 24.0 ft, the flexural buckling critical stress was Fcr = 29.1 ksi. By Section E7.1, at Fcr = 29.1 ksi, the wide wall effective width must be determined but the narrow wall is fully effective. Thus, the axial strength is reduced because of local buckling of the wide wall. Table E.10 repeats the example analysis for two other column effective lengths and compares those results to the results for Lc = 24 ft calculated previously. For Lc = 18.0 ft, the flexural buckling critical stress, Fcr = 36.9 ksi, is high enough that both the wide and narrow walls must have their effective width determined according to Equation E7-3. For Lc = 40.0 ft the flexural buckling critical stress, Fcr = 12.2 ksi, is low enough that there will be no local buckling of either wall and the actual widths will be used according to Equation E7-2.

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Table E.10. Analysis of HSS128x Column at Different Effective Lengths Effective length, Lc (ft) Check Table B4.1 criterion (same as for Lc = 24.0 ft). r  (narrow wall) = 43.0 > r  (wide wall) = 66.0 > r Fcr (ksi)

18.0

24.0

40.0

33.7 Yes Yes

33.7 Yes Yes

33.7 Yes Yes

36.9

29.1

12.2

39.2    43.0

44.2    43.0

68.2    43.0

Yes

No

No

58.5 7.05

– –

– –

39.2    66.0

44.2    66.0

68.2    66.0

Yes

Yes

No

24.8 7.88

24.8 8.66

– –

Effective area, Ae (in.2) Compressive strength Pn (kips) LRFD, c Pn (kips)

5.35

5.77

6.76

197 177

168 151

82.5 74.2

ASD, Pn c (kips)

118

101

49.4

Check AISC Specification Section E7.1 criteria. Narrow wall:

r

Fy Fcr

Local buckling reduction per AISC Specification Section E7.1? Fel (ksi) be (in.) Wide wall:

r

Fy Fcr

Local buckling reduction per AISC Specification Section E7.1? Fel (ksi) he (in.)

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EXAMPLE E.11 PIPE COMPRESSION MEMBER Given: Select an ASTM A53 Grade B Pipe compression member with a length of 30 ft to support a dead load of 35 kips and live load of 105 kips in axial compression. The column is pin-connected at the ends in both axes and braced at the midpoint in the y-y direction. The solution will be provided using: (1) AISC Manual Tables (2) Calculations using AISC Specification provisions

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A53 Grade B Fy = 35 ksi Fu = 60 ksi From ASCE/SEI 7, Chapter 2, the required compressive strength is: ASD

LRFD Pu  1.2  35 kips   1.6 105 kips 

Pa  35 kips  105 kips  140 kips

 210 kips (1) AISC Manual Table Solution

From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0. Therefore, Lcx = KxLx = 1.0(30 ft) = 30.0 ft and Lcy = KyLy = 1.0(15 ft) = 15.0 ft. Buckling about the x-x axis controls. Enter AISC Manual Table 4-6 with Lc = 30.0 ft and select the lightest section with sufficient available strength to support the required strength. Try a 10-in. Standard Pipe. From AISC Manual Table 4-6, the available strength in axial compression is: ASD

LRFD

Pn  148 kips  140 kips o.k. c

c Pn  222 kips  210 kips o.k.

Available strength can also be determined by hand calculations, as demonstrated in the following. (2) Calculation Using AISC Specification Provisions

From AISC Manual Table 1-14, the geometric properties are as follows: Pipe 10 Std.

Ag = 11.5 in.2 r = 3.68 in. D  31.6   = t

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No Pipes shown in AISC Manual Table 4-6 are slender at 35 ksi, so no local buckling check is required; however, some round HSS are slender at higher steel strengths. The following calculations illustrate the required check. Limiting Width-to-Thickness Ratio Determine the wall limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 9:  r  0.11

E Fy

 29, 000 ksi   0.11   35 ksi   91.1    r ; therefore, the pipe is not slender

Critical Stress, Fcr Lc  30.0 ft 12 in./ft   3.68 in. r  97.8

4.71

E 29, 000 ksi  4.71 Fy 35 ksi  136  97.8, therefore, use AISC Specification Equation E3-2

Fe 



2 E  Lc     r 

(Spec. Eq. E3-4)

2

2  29, 000 ksi 

 97.8 2

 29.9 ksi Fy   Fcr   0.658 Fe  Fy     35 ksi         0.658 29.9 ksi    35 ksi       21.4 ksi

(Spec. Eq. E3-2)

Available Compressive Strength Pn  Fcr Ag

(Spec. Eq. E3-1)



  21.4 ksi  11.5 in.2



 246 kips

From AISC Specification Section E1, the available compressive strength is:

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ASD

LRFD c = 0.90

c = 1.67

c Pn  0.90  246 kips 

Pn 246 kips  c 1.67  147 kips  140 kips

 221 kips  210 kips o.k.

Note that the design procedure would be similar for a round HSS column.

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EXAMPLE E.12 BUILT-UP I-SHAPED MEMBER WITH DIFFERENT FLANGE SIZES Given: Compute the available strength of a built-up compression member with a length of 14 ft, as shown in Figure E.12-1. The ends are pinned. The outside flange is PLw in. 5 in., the inside flange is PLw in. 8 in., and the web is PLa in. 102 in. The material is ASTM A572 Grade 50.

Fig. E.12-1. Column geometry for Example E.12.

Solution: From AISC Manual Table 2-5, the material properties are as follows: ASTM A572 Grade 50 Fy = 50 ksi Fu = 65 ksi There are no tables for special built-up shapes; therefore, the available strength is calculated as follows. Slenderness Check Check outside flange slenderness. From AISC Specification Table B4.1a note [a], calculate kc. kc  =

4 h tw 4

102 in. a in.  0.756, 0.35  kc  0.76

o.k.

For the outside flange, the slenderness ratio is:

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b t 2.50 in.  w in.  3.33



Determine the limiting slenderness ratio, r, from AISC Specification Table B4.1a, Case 2:  r  0.64  0.64

kc E Fy 0.756  29, 000 ksi  50 ksi

 13.4    r ; therefore, the outside flange is not slender

Check inside flange slenderness. b t 4.00 in.  w in.  5.33



   r ; therefore, the inside flange is not slender

Check web slenderness. h t 102 in.  a in.  28.0



Determine the limiting slenderness ratio, r, for the web from AISC Specification Table B4.1a, Case 5:  r  1.49  1.49

E Fy 29, 000 ksi 50 ksi

 35.9    r ; therefore, the web is not slender

Section Properties (ignoring welds)

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Ag  b fi t fi  htw  b fo t fo   8.00 in. w in.  102 in. a in.   5.00 in. w in.  13.7 in.2 y

Ai yi Ai

 6.00 in.  11.6 in.   3.94 in.   6.00 in.   3.75 in.   0.375 in.  2

2

2

6.00 in.2  3.94 in.2  3.75 in.2

 6.91 in.

Note that the center of gravity about the x-axis is measured from the bottom of the outside flange.  bh3  I x     Ad 2   12    8.00 in. w in.3    a in.102 in.3  2 2    8.00 in. w in. 4.72 in.      a in.102 in. 0.910 in.  12 12       5.00 in. w in.3  2    5.00 in. w in. 6.54 in.  12    334 in.4 rx  

Ix A 334 in.4

13.7 in.2  4.94 in. Iy   

bh3 12

 w in.8.00 in.3 102 in. a in.3  w in. 5.00 in.3 12





12

12

4

 39.9 in.

ry  

Iy A 39.9 in.4

13.7 in.2  1.71 in.

Elastic Buckling Stress about the x-x Axis From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0. Therefore, Lcx = Lcy = Lcz = KL = 1.0(14 ft) = 14.0 ft. The effective slenderness ratio about the x-axis is:

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Lcx 14.0 ft 12 in./ft   4.94 in. rx  34.0

Fe 



2 E  Lc     r 

(Spec. Eq. E3-4)

2

2  29, 000 ksi 

 34.0 2

 248 ksi

does not control

Flexural-Torsional Elastic Buckling Stress Calculate the torsional constant, J, using AISC Design Guide 9, Equation 3.4:

J  

bt 3 3

8.00 in. w in.3 102 in. a in.3  5.00 in. w in.3 3





3

3

4

 2.01 in.

Distance between flange centroids: ho  d 

t fi 2



t fo

2 w in. w in.  12.0 in.   2 2  11.3 in.

Warping constant: Cw  

t f ho 2  b fi 3b fo3    12  b fi 3  b fo3 

 w in.11.3 in.2  8.00 in.3  5.00 in.3  12   8.00 in.3   5.00 in.3 

 802 in.6

Due to symmetry, both the centroid and the shear center lie on the y-axis. Therefore, xo  0. The distance from the center of the outside flange to the shear center is:  b fi 3  e  ho  3  3  b fi  b fo      8.00 in.3    11.3 in.   8.00 in.3   5.00 in.3   9.08 in.

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Add one-half the flange thickness to determine the shear center location measured from the bottom of the outside flange. e

tf w in.  9.08 in.  2 2  9.46 in.

tf   yo   e    y 2   9.46 in.  6.91 in.  2.55 in. ro 2  xo2  yo2 

Ix  I y Ag

(Spec. Eq. E4-9)

  0   (2.55 in.) 2  2

334 in.4  39.9 in.4 13.7 in.2

 33.8 in.2

H  1  1

xo2  yo2

(Spec. Eq. E4-8)

ro 2

 0 2   2.55 in.2 33.8 in.2

 0.808

The effective slenderness ratio about the y-axis is: Lcy ry



14.0 ft 12 in./ft  1.71 in.

 98.2

Fey 



2 E  Lcy     ry 

(Spec. Eq. E4-6)

2

2  29, 000 ksi 

 98.2 2

 29.7 ksi  2 ECw  1  GJ  Fez   2 2  Lcz  Ag ro



(Spec. Eq. E4-7)



 2  29, 000 ksi  802 in.6    11, 200 ksi  2.01 in.4 2  14.0 ft 12 in./ft  





 1   2 2    13.7 in. 33.8 in.





 66.2 ksi

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   

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 4 Fey Fez H   1  1  2   Fey  Fez  

 Fey  Fez Fe    2H

   

(Spec. Eq. E4-3)

 29.7 ksi  66.2 ksi   4  29.7 ksi  66.2 ksi  0.808      1  1  2  0.808   29.7 ksi  66.2 ksi 2      26.4 ksi

controls

Torsional and flexural-torsional buckling governs. Fy 50 ksi  Fe 26.4 ksi  1.89

Fy  2.25 : Fe

Because

Fy  Fcr   0.658 Fe  



  Fy  

(Spec. Eq. E3-2)



 0.6581.89  50 ksi   22.7 ksi

Available Compressive Strength Pn  Fcr Ag

(Spec. Eq. E3-1)



  22.7 ksi  13.7 in.2



 311 kips

From AISC Specification Section E1, the available compressive strength is: ASD

LRFD c = 0.90

c = 1.67

c Pn  0.90  311 kips 

Pn 311 kips  c 1.67  186 kips

 280 kips

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EXAMPLE E.13 DOUBLE-WT COMPRESSION MEMBER Given:

Determine the available compressive strength for an ASTM A992 double-WT920 compression member, as shown in Figure E.13-1. Assume that 2-in.-thick connectors are welded in position at the ends and at equal intervals, “a”, along the length. Use the minimum number of intermediate connectors needed to force the two WT-shapes to act as a single built-up compression member.

Fig. E.13-1. Double-WT compression member in Example E.13. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Tee ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-8 the geometric properties for a single WT920 are as follows: A = 5.88 in.2 d = 8.95 in. tw = 0.315 in. d/tw = 28.4 Ix = 44.8 in.4 Iy = 9.55 in.4 rx = 2.76 in. ry = 1.27 in. y = 2.29 in. J = 0.404 in.4 Cw = 0.788 in.6 From mechanics of materials, the combined section properties for two WT920’s, flange-to-flange, spaced 2-in. apart, are as follows: A  Asingle tee



 2 5.88 in.2



 11.8 in.2

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



I x   I x  Ay 2



 2  44.8 in.4  5.88 in.2 

  2.29 in.  4 in.  2

 165 in.4 Ix A

rx 

165 in.4



11.8 in.2  3.74 in.

I y  I y



single tee

 2 9.55 in.4



 19.1 in.4

Iy A

ry  

19.1 in.4

11.8 in.2  1.27 in. J  J single tee



 2 0.404 in.4



 0.808 in.4 For the double-WT (cruciform) shape shown in Figure E.13-2 it is reasonable to take Cw  0 and ignore any warping contribution to column strength.

Fig. E.13-2. Double-WT shape cross section.

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The y-axis of the combined section is the same as the y-axis of the single section. When buckling occurs about the yaxis, there is no relative slip between the two WTs. For buckling about the x-axis of the combined section, the WTs will slip relative to each other unless restrained by welded or slip-critical end connections. Intermediate Connectors Dimensional Requirements Determine the minimum number of intermediate connectors required. From AISC Specification Section E6.2, the maximum slenderness ratio of each tee should not exceed three-fourths times the maximum slenderness ratio of the double-WT built-up section. For a WT920, the minimum radius of gyration is: ri  ry  1.27 in.

Use K = 1.0 for both the single tee and the double tee; therefore, Lcy = KyLy = 1.0(9 ft) = 9.00 ft: 3  Lcy  a      r  i  single tee 4  rmin double tee

a

3  ry  single tee

4  ry double tee

 Lcy double tee

3  1.27 in.     9.00 ft 12 in./ft   4  1.27 in.    81.0 in. 

Thus, one intermediate connector at mid-length [a = (4.5 ft)(12 in./ft) = 54.0 in.] satisfies AISC Specification Section E6.2 as shown in Figure E.13-3.

Figure E.13-3. Minimum connectors required for double-WT compression member.

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Flexural Buckling and Torsional Buckling Strength For the WT920, the stem is slender because d/tw = 28.4 > 0.75 29, 000 ksi 50 ksi = 18.1 (from AISC Specification Table B4.1a, Case 4). Therefore, the member is a slender element member and the provisions of Section E7 are followed. Determine the elastic buckling stress for flexural buckling about the y- and x-axes, and torsional buckling. Then, determine the effective area considering local buckling, the critical buckling stress, and the nominal strength. Elastic Buckling Stress about the y-y Axis Lcy  9.00 ft 12 in./ft   1.27 in. ry  85.0

Fey 



2 E  Lcy     ry 

(Spec. Eq. E4-6)

2

2  29, 000 ksi 

 85.0 2

 39.6 ksi

controls

Elastic Buckling Stress about the x-x Axis Flexural buckling about the x-axis is determined using the modified slenderness ratio to account for shear deformation of the intermediate connectors. Note that the provisions of AISC Specification Section E6.1, which require that Lc r be replaced with  Lc r m , apply if “the buckling mode involves relative deformations that produce shear forces in the connectors between individual shapes…”. Relative slip between the two sections occurs for buckling about the x-axis so the provisions of the section apply only to buckling about the x-axis. The connectors are welded at the ends and the intermediate point. The modified slenderness is calculated using the spacing between intermediate connectors: a   4.5 ft 12.0 in./ft   54.0 in. ri  ry  1.27 in.

a 54.0 in.  ri 1.27 in.  42.5

Because a ri  40, use AISC Specification Equation E6-2b. 2

 Lc   Lc   Ki a   r    r   r   m  o  i 

2

(Spec. Eq. E6-2b)

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where Lcx  Lc     rx  r o 

 9.00 ft 12 in./ft  3.74 in.

 28.9 K i a 0.86  4.50 ft 12 in./ft   1.27 in. ri  36.6

Thus,  Lc  2 2  r    28.9    36.6   m  46.6

Fex 



2 E  Lcx   r   x 

(Spec. Eq. E4-5)

2

2  29, 000 ksi 

 46.6 2

 132 ksi

Torsional Buckling Elastic Stress  2 ECw  1  GJ  Fe   2  Lcz  Ix  I y

(Spec. Eq. E4-2)

The cruciform section made up of two back-to-back WT's has virtually no warping resistance, thus the warping contribution is ignored and Specification Equation E4-2 becomes:

Fe  

GJ Ix  I y

11, 200 ksi   0.808 in.4 

165 in.4  19.1 in.4  49.2 ksi Critical Stress Use the smallest elastic buckling stress, Fe, from the limit states considered above to determine Fcr by AISC Specification Equation E3-2 or Equation E3-3, as follows: Fe  39.6 ksi

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Fy 50 ksi  Fe 39.6 ksi  1.26 Fy  2.25, Fe

Because

Fy   Fcr  0.658 Fe  



  Fy  

(Spec. Eq. E3-2)



 0.6581.26  50 ksi   29.5 ksi

Effective Area Since the stem was previously shown to be slender, calculate the limits of AISC Specification Section E7.1 to determine if the stem is fully effective or if there is a reduction in effective area due to local buckling of the stem.   28.4

 r  0.75  0.75

E Fy 29, 000 ksi 50 ksi

 18.1

r

Fy Fcr

 18.1

50 ksi 29.5 ksi

 23.6

Because    r Fy Fcr , the stem will not be fully effective and there will be a reduction in effective area due to local buckling of the stem. The effective width imperfection adjustment factors can be determined from AISC Specification Table E7.1, Case (c), as follows. c1  0.22 c2  1.49

Determine the elastic local buckling stress from AISC Specification Section E7.1. 2

   Fel   c2 r  Fy      18.1    1.49    28.4     45.1 ksi

(Spec. Eq. E7-5) 2

 50 ksi 

Determine the effective width of the tee stem and the resulting effective area, where b = d = 8.95 in.

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 F  F be  b 1  c1 el  el Fcr  Fcr   45.1 ksi  45.1 ksi   8.95 in. 1  0.22  29.5 ksi  29.5 ksi   8.06 in.

(Spec. Eq. E7-3)

Ae   A   tw  b  be  





  2  5.88 in.2   2  0.315 in. 8.95 in.  8.06 in. 2

 11.2 in.

Available Compressive Strength Pn  Fcr Ae



  29.5 ksi  11.2 in.2

(Spec. Eq. E7-1)



 330 kips

From AISC Specification Section E1, the available compressive strength is: LRFD

ASD

c  0.90

c  1.67

c Pn  0.90  330 kips 

Pn 330 kips  c 1.67  198 kips

 297 kips

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EXAMPLE E.14 ECCENTRICALLY LOADED SINGLE-ANGLE COMPRESSION MEMBER (LONG LEG ATTACHED) Given: Determine the available strength of an eccentrically loaded ASTM A36 L842 single angle compression member, as shown in Figure E.14-1, with an effective length of 5 ft. The long leg of the angle is the attached leg, and the eccentric load is applied at 0.75t as shown. Use the provisions of the AISC Specification and compare the results to the available strength found in AISC Manual Table 4-12.

Fig. E.14-1. Eccentrically loaded single-angle compression member in Example E.14.

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-7: L842 x = 0.854 in.

y = 2.84 in. A = 5.80 in.2 Ix = 38.6 in.4 Iy = 6.75 in.4 Iz = 4.32 in.4 rz = 0.863 in. tan  = 0.266 From AISC Shapes Database V15.0: Iw SwA SwB SwC SzA SzB SwC

= 41.0 in.4 = 12.4 in.3 = 16.3 in.3 = 7.98 in.3 = 1.82 in.3 = 2.77 in.3 = 5.81 in.3

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Fig. E.14-2. Geometry about principal axes. The load is applied at the location shown in Figure E.14-2. Determine the eccentricities about the major (w-w axis) and minor (z-z axis) principal axes for the load, P. From AISC Manual Table 1-7, the angle of the principal axes is found to be α = tan1(0.266) = 14.9°. Using the geometry shown in Figures E.14-2 and E.14-3:  0.5b  y  ew   x  0.75t    0.5b  y  tan   sin      cos    0.5  8.00 in.  2.84 in.   0.854 in.  0.75 2 in.   0.5  8.00 in.  2.84 in.  0.266   sin14.9      cos14.9   





 1.44 in.

ez   x  0.75t  cos    0.5b  y  sin   0.854 in.  0.75 2 in.   cos14.9   0.5  8.00 in.  2.84 in.  sin14.9   0.889 in. Because of these eccentricities, the moment resultant has components about both principal axes; therefore, the combined stress provisions of AISC Specification Section H2 must be followed. f ra f rbw f rbz   Fca Fcbw Fcbz

 1.0

(Spec. Eq. H2-1)

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Fig. E.14-3. Applied moments and eccentric axial load. Due to the load and the given eccentricities, moments about the w-w and z-z axes will have different effects on points A, B and C. The axial force will produce a compressive stress and the moments, where positive moments are in the direction shown in Figure E.14-3, will produce stresses with a sign indicated by the sense given in the following. In this example, compressive stresses will be taken as positive and tensile stresses will be taken as negative. Point A B C

Caused by Mw tension tension compression

Caused by Mz tension compression tension

Available Compressive Strength Check the slenderness of the longest leg for uniform compression. b t 8.00 in.  2 in.  16.0



Check the slenderness of the shorter leg for uniform compression. d t 4.00 in.  2 in.  8.00



From AISC Specification Table B4.1a, Case 3, the limiting width-to-thickness ratio is:  r  0.45  0.45

E Fy 29, 000 ksi 36 ksi

 12.8

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Because b/t = 16.0 > 12.8, the longer leg is classified as a slender element for compression. Because d/t = 8.00 < 12.8, the shorter leg is classified as a nonslender element for compression. Determine if torsional and flexural-torsional buckling is applicable, using the provisions of AISC Specification Section E4.   16.0

E 29, 000 ksi  0.71 36 ksi Fy

0.71

 20.2

Because   0.71 E / Fy , torsional and flexural-torsional buckling is not applicable. Determine the critical stress, Fcr , with Lc = (5.00 ft)(12 in./ft) = 60.0 in. for buckling about the z-z axis. Lcz 60.0 in.  rz 0.863 in.  69.5 Fe 



2 E  Lcz   r   z 

(Spec. Eq. E3-4)

2

2  29, 000 ksi 

 69.5 2

 59.3 ksi Fy 36 ksi  Fe 59.3 ksi  0.607 Fy  2.25 : Fe

Because

Fy   Fcr  0.658 Fe  



  Fy  

 0.6580.607

(Spec. Eq. E3-2)

 36 ksi 

 27.9 ksi

Because the longer leg was found to be slender, the limits of AISC Specification Section E7.1 must be evaluated to determine if the leg is fully effective for compression or if a reduction in effective area must be taken to account for local buckling in the longer leg.   16.0

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r

Fy Fcr

36 ksi 27.9 ksi

 12.8  14.5

Because   14.5, there will be a reduction in effective area due to local buckling in the longer leg. Determine the effective width imperfection adjustment factors per AISC Specification Table E7.1 as follows. c1  0.22 c2  1.49

Determine the elastic local buckling stress from AISC Specification Section E7.1. 2

   Fel   c2 r  Fy      12.8    1.49    16.0     51.2 ksi

(Spec. Eq. E7-5) 2

 36 ksi 

Determine the effective width of the angle leg and the resulting effective area.  F  F be  b 1  c1 el  el F cr  Fcr   51.2 ksi  51.2 ksi   8.00 in. 1  0.22  27.9 ksi  27.9 ksi   7.61 in.

(Spec. Eq. E7-3)

Ae  Ag  t   b  be   5.80 in.2  2 in. 8.00 in.  7.61 in.  5.61 in.2

Available Compressive Strength Pn  Fcr Ae



  27.9 ksi  5.61 in.2

(Spec. Eq. E7-1)



 157 kips

From AISC Specification Section E1, the available compressive strength is: LRFD

ASD

c  0.90

c  1.67

c Pn  0.90 157 kips 

Pn 157 kips  c 1.67  94.0 kips

 141 kips

Determine the available flexural strengths, Mcbw and Mcbz, and the available flexural stresses at each point on the cross section.

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Yielding Consider the limit state of yielding for bending about the w-w and z-z axes at points A, B and C, according to AISC Specification Section F10.1. w-w axis: M ywA  Fy S wA





  36 ksi  12.4 in.3  446 kip-in.

M nwA  1.5M ywA

(from Spec. Eq. F10-1)

 1.5  446 kip-in.  669 kip-in. M ywB  Fy S wB



  36 ksi  16.3 in.3



 587 kip-in.

M nwB  1.5M ywB

(from Spec. Eq. F10-1)

 1.5  587 kip-in.  881 kip-in. M ywC  Fy S wC



  36 ksi  7.98 in.3



 287 kip-in.

M nwC  1.5M ywC

(from Spec. Eq. F10-1)

 1.5  287 kip-in.  431 kip-in. z-z axis: M yzA  Fy S zA



  36 ksi  1.82 in.3



 65.5 kip-in.

M nzA  1.5M yzA

(from Spec. Eq. F10-1)

 1.5  65.5 kip-in.  98.3 kip-in.

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M yzB  Fy S zB



  36 ksi  2.77 in.3



 99.7 kip-in.

M nzB  1.5M yzB

(from Spec. Eq. F10-1)

 1.5  99.7 kip-in.  150 kip-in. M yzC  Fy S zC



  36 ksi  5.81 in.3



 209 kip-in.

M nzC  1.5M yzC  1.5  209 kip-in.

(from Spec. Eq. F10-1)

 314 kip-in. Select the least Mn for each axis. For the limit state of yielding about the w-w axis: M nw  431 kip-in. at point C

For the limit state of yielding about the z-z axis: M nz  98.3 kip-in. at point A

Lateral-Torsional Buckling From AISC Specification Section F10.2, the limit state of lateral-torsional buckling of a single angle without continuous restraint along its length is a function of the elastic lateral-torsional buckling moment about the major principal axis. For bending about the major principal axis for a single angle: M cr 

2    r   r 9 EArz tCb   1   4.4 w z   4.4 w z  8 Lb  Lbt  Lbt    

(Spec. Eq. F10-4)

From AISC Specification Section F1, for uniform moment along the member length, Cb = 1.0. From AISC Specification Commentary Table C-F10.1, an L842 has w = 5.48 in. From AISC Specification Commentary Figure C-F10.4b, with the tip of the long leg (point C) in compression for bending about the w-axis, w is taken as negative. Thus: M cr 





9  29, 000 ksi  5.80 in.2  0.863 in.2 in.1.0  8  60.0 in.

2     5.48 in. 0.863 in.   5.48 in. 0.863 in.     1   4.4  4.4    60.0 in.2 in.   60.0 in.2 in.       712 kip-in.

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M ywC 287 kip-in.  712 kip-in. M cr  0.403

Because M ywC M cr  1.0, determine Mn as follows:

 M ywC M nwC  1.92  1.17  M cr 





  M ywC  1.5M ywC 

(from Spec. Eq. F10-2)

 1.92  1.17 0.403  287 kip-in.  1.5  287 kip-in.  338 kip-in.  431 kip-in.  338 kip-in. Leg Local Buckling From AISC Specification Section F10.3, the limit state of leg local buckling applies when the toe of the leg is in compression. As discussed previously and indicated in Table E.14-1, the only case in which a toe is in compression is point C for bending about the w-w axis. Thus, determine the slenderness of the long leg as a compression element subject to flexure. From AISC Specification Table B4.1b, Case 12:  p  0.54  0.54

E Fy 29, 000 ksi 36 ksi

 15.3  r  0.91  0.91

E Fy 29, 000 ksi 36 ksi

 25.8 b t 8.0 in.  2 in.



 16.0

Because  p     r , the angle is noncompact for flexure for this loading. From AISC Specification Equation F106:

  b  Fy  M nwC  Fy S wC  2.43  1.72      t  E    36 ksi    36 ksi  7.98 in.3  2.43  1.72 16.0   29, 000 ksi    420 kip-in.





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(from Spec. Eq. F10-6)

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Table E.14-1 provides a summary of nominal flexural strength at each point. T indicates the point is in tension and C indicates it is in compression. Table E.14-1 Yielding Lateral-Torsional Buckling Point Mnw, kip-in. Mnz, kip-in. Mnw, kip-in. Mnz, kip-in. A 669 T 98.3 T   B 881 T 150 C   C 431 C 314 T 338 C  Note: () indicates that the limit state is not applicable to this point.

Leg Local Buckling Mnw, kip-in. Mnz, kip-in.     420 C 

Available Flexural Strength Select the controlling nominal flexural strength for the w-w and z-z axes. For the w-w axis: M nw  338 kip-in.

For the z-z axis: M nz  98.3 kip-in.

From AISC Specification Section F1, determine the available flexural strength for each axis, w-w and z-z, as follows: LRFD b  0.90

M cbw  b M nw  0.90  338 kip-in.  304 kip-in.

M cbz  b M nz  0.90  98.3 kip-in.  88.5 kip-in.

ASD

b  1.67 M nw b 338 kip-in.  1.67  202 kip-in.

M cbw 

M nz b 98.3 kip-in.  1.67  58.9 kip-in.

M cbz 

Required Flexural Strength The load on the column is applied at eccentricities about the w-w and z-z axes resulting in the following moments: M w  Pr ew  Pr 1.44 in. and

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M z  Pr ez  Pr  0.889 in. The combination of axial load and moment will produce second-order effects in the column which must be accounted for. Using AISC Specification Appendix 8.2, an approximate second-order analysis can be performed. The required second-order flexural strengths will be B1w Mw and B1z Mz, respectively, where B1 

Cm  1.0 P 1 r Pe1

(Spec. Eq. A-8-3)

and   1.0 (LRFD)   1.6 (ASD) Cm = 1.0 for a column with uniform moment along its length For each axis, parameters Pe1w and Pe1z , as used in the moment magnification terms, B1w and B1z , are: Pe1w 



2 EI w

(from Spec. Eq. A-8-5)

 Lc1 2 2  29, 000 ksi   41.0 in.4   60.0 in.2

 3, 260 kips Pe1z  

2 EI z

(from Spec. Eq. A-8-5)

 Lc1 2 2 (29, 000 ksi)(4.32 in.4 )

 60.0 in.2

 343 kips

and Cm P 1 r Pe1w 1.0  Pr 1 3, 260 kips

B1w 

(Spec. Eq. A-8-3)

Cm P 1 r Pe1z 1.0  Pr 1 343 kips

B1z 

(Spec. Eq. A-8-3)

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Thus, the required second-order flexural strengths are: 1.0    Pr M rw  Pr 1.44 in.   1  3, 260 kips   

M rz

1.0    Pr   Pr  0.889 in.  1    343 kips 

Interaction of Axial and Flexural Strength Evaluate the interaction of axial and flexural stresses according to the provisions of AISC Specification Section H2. The interaction equation is given as: f ra f rbw f rbz   Fca Fcbw Fcbz

 1.0

(Spec. Eq. H2-1)

where the stresses are to be considered at each point on the cross section with the appropriate sign representing the sense of the stress. Because the required stress and available stress at any point are both functions of the same section property, A or S, it is possible to convert Equation H2-1 from a stress based equation to a force based equation where the section properties will cancel. Substituting the available strengths and the expressions for the required second-order flexural strengths into AISC Specification Equation H2-1 yields: LRFD 1.0   Pu 1.44 in.  Pu  P 1.0  u  141 kips 304 kip-in.  1   3, 260 kips  1    Pu  0.889 in.    1.0 P  u   88.5 kip-in.   1  343 kips   

ASD

 1.0

Pa 1.44 in.  Pa 1.0     94.0 kips 202 kip-in. 1  1.6 Pa    3, 260 kips   1.0  Pa  0.889 in.   1      58.9 kip-in.   1  1.6 Pa   343 kips   

These interaction equations must now be applied at each critical point on the section, points A, B and C using the appropriate sign for the sense of the resulting stress, with compression taken as positive. For point A, the w term is negative and the z term is negative. Thus: LRFD 1.0   Pu 1.44 in.  Pu 1.0 Pu    141 kips 304 kip-in.  1   3, 260 kips  1    Pu  0.889 in.   1.0 Pu     88.5 kip-in.   1  343 kips   

By iteration, Pu = 88.4 kips.

 1.0

ASD P 1.44 in.   Pa 1.0 a    94.0 kips 202 kip-in.  1  1.6 Pa     3, 260 kips   1.0  Pa  0.889 in.   1      58.9 kip-in.   1  1.6 Pa   343 kips    By iteration, Pa = 57.7 kips.

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For point B, the w term is negative and the z term is positive. Thus: LRFD

ASD

1.0   Pu 1.44 in.  Pu 1.0 Pu    141 kips 304 kip-in.  1   3, 260 kips   1.0 1    Pu  0.889 in.       1  1.0 Pu  88.5 kip-in.    343 kips   

By iteration, Pu = 67.7 kips.

Pa 1.44 in.  Pa 1.0     1.6 Pa 94.0 kips 202 kip-in. 1     3, 260 kips   1.0  Pa  0.889 in.   1      58.9 kip-in.   1  1.6 Pa   343 kips   

By iteration, Pa = 44.6 kips.

For point C, the w term is positive and the z term is negative. Thus: LRFD

ASD

1.0   Pu 1.44 in.  Pu 1.0 Pu    141 kips 304 kip-in.  1   3, 260 kips   1.0 1    Pu  0.889 in.       1  1.0 Pu  88.5 kip-in.    343 kips   

By iteration, Pu = 156 kips.

Pa 1.44 in.  Pa 1.0     1.6 P 94.0 kips 202 kip-in. 1 a     3, 260 kips   1.0  Pa  0.889 in.   1      58.9 kip-in.   1  1.6 Pa   343 kips   

By iteration, Pa = 99.5 kips.

Governing Available Strength LRFD From the above iterations,

From the above iterations,

ASD

Pu = 67.7 kips

Pa = 44.6 kips

From AISC Manual Table 4-12,

From AISC Manual Table 4-12,

Pn  67.7 kips

Pn  44.6 kips 

Thus, the calculations demonstrate how the values for this member in AISC Manual Table 4-12 can be confirmed.

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Chapter F Design of Members for Flexure INTRODUCTION This Specification chapter contains provisions for calculating the flexural strength of members subject to simple bending about one principal axis. Included are specific provisions for I-shaped members, channels, HSS, box sections, tees, double angles, single angles, rectangular bars, rounds and unsymmetrical shapes. Also included is a section with proportioning requirements for beams and girders. There are selection tables in the AISC Manual for standard beams in the commonly available yield strengths. The section property tables for most cross sections provide information that can be used to conveniently identify noncompact and slender element sections. LRFD and ASD information is presented side-by-side. Most of the formulas from this chapter are illustrated by the following examples. The design and selection techniques illustrated in the examples for both LRFD and ASD will result in similar designs. F1. GENERAL PROVISIONS Selection and evaluation of all members is based on deflection requirements and strength, which is determined as the design flexural strength, bMn, or the allowable flexural strength, Mn/b, where Mn = the lowest nominal flexural strength based on the limit states of yielding, lateral torsional-buckling, and local buckling, where applicable b = 0.90 (LRFD) b = 1.67 (ASD) This design approach is followed in all examples. The term Lb is used throughout this chapter to describe the length between points which are either braced against lateral displacement of the compression flange or braced against twist of the cross section. Requirements for bracing systems and the required strength and stiffness at brace points are given in AISC Specification Appendix 6. The use of Cb is illustrated in several of the following examples. AISC Manual Table 3-1 provides tabulated Cb values for some common situations. F2. DOUBLY SYMMETRIC COMPACT I-SHAPED MEMBERS AND CHANNELS BENT ABOUT THEIR MAJOR AXIS AISC Specification Section F2 applies to the design of compact beams and channels. As indicated in the User Note in Section F2 of the AISC Specification, the vast majority of rolled I-shaped beams and channels fall into this category. The curve presented as a solid line in Figure F-1 is a generic plot of the nominal flexural strength, Mn, as a function of the unbraced length, Lb. The horizontal segment of the curve at the far left, between Lb = 0 ft and Lp, is the range where the strength is limited by flexural yielding. In this region, the nominal strength is taken as the full plastic moment strength of the section as given by AISC Specification Equation F2-1. In the range of the curve at the far right, starting at Lr, the strength is limited by elastic buckling. The strength in this region is given by AISC Specification Equation F2-3. Between these regions, within the linear region of the curve between Mn = Mp at Lp on the left, and Mn = 0.7My = 0.7FySx at Lr on the right, the strength is limited by inelastic buckling. The strength in this region is provided in AISC Specification Equation F2-2. The curve plotted as a heavy solid line represents the case where Cb = 1.0, while the heavy dashed line represents the case where Cb exceeds 1.0. The nominal strengths calculated in both AISC Specification Equations F2-2 and F2-3 are linearly proportional to Cb, but are limited to Mp as shown in the figure.

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Fig. F-1. Nominal flexural strength versus unbraced length. M n  M p  Fy Z x

  Lb  L p M n  Cb  M p   M p  0.7 Fy S x     Lr  L p

(Spec. Eq. F2-1)

    M p  

M n  Fcr S x  M p

(Spec. Eq. F2-2) (Spec. Eq. F2-3)

where Fcr 

Cb 2 E  Lb  r   ts 

2

1  0.078

Jc  Lb    S x ho  rts 

2

(Spec. Eq. F2-4)

The provisions of this section are illustrated in Example F.1 (W-shape beam) and Example F.2 (channel). Inelastic design provisions are given in AISC Specification Appendix 1. Lpd, the maximum unbraced length for prismatic member segments containing plastic hinges is less than Lp. F3. DOUBLY SYMMETRIC I-SHAPED MEMBERS WITH COMPACT WEBS AND NONCOMPACT OR SLENDER FLANGES BENT ABOUT THEIR MAJOR AXIS

The strength of shapes designed according to this section is limited by local buckling of the compression flange. Only a few standard wide-flange shapes have noncompact flanges. For these sections, the strength reduction for Fy = 50 ksi steel varies. The approximate percentages of Mp about the strong axis that can be developed by noncompact members when braced such that Lb  Lp are shown as follows: W2148 = 99% W1012 = 99% W68.5 = 97%

W1499 = 99% W831 = 99%

W1490 = 97% W810 = 99%

W1265 = 98% W615 = 94%

The strength curve for the flange local buckling limit state, shown in Figure F-2, is similar in nature to that of the lateral-torsional buckling curve. The horizontal axis parameter is = bf /2tf. The flat portion of the curve to the left of pf is the plastic yielding strength, Mp. The curved portion to the right of rf is the strength limited by elastic

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buckling of the flange. The linear transition between these two regions is the strength limited by inelastic flange buckling.

Fig. F-2. Flange local buckling strength. M n  M p  Fy Z x

    pf M n  M p   M p  0.7 Fy S x     rf   pf Mn 

(Spec. Eq. F2-1)   

0.9 Ekc S x 2

(Spec. Eq. F3-1)

(Spec. Eq. F3-2)

where kc 

4 h tw

and shall not be taken less than 0.35 nor greater than 0.76 for calculation purposes.

The strength reductions due to flange local buckling of the few standard rolled shapes with noncompact flanges are incorporated into the design tables in Part 3 and Part 6 of the AISC Manual. There are no standard I-shaped members with slender flanges. The noncompact flange provisions of this section are illustrated in Example F.3. F4. OTHER I-SHAPED MEMBERS WITH COMPACT OR NONCOMPACT WEBS BENT ABOUT THEIR MAJOR AXIS

This section of the AISC Specification applies to doubly symmetric I-shaped members with noncompact webs and singly symmetric I-shaped members (those having different flanges) with compact or noncompact webs. F5. DOUBLY SYMMETRIC AND SINGLY SYMMETRIC I-SHAPED MEMBERS WITH SLENDER WEBS BENT ABOUT THEIR MAJOR AXIS

This section applies to doubly symmetric and singly symmetric I-shaped members with slender webs, formerly designated as “plate girders”.

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F6. I-SHAPED MEMBERS AND CHANNELS BENT ABOUT THEIR MINOR AXIS

I-shaped members and channels bent about their minor axis are not subject to lateral-torsional buckling. Rolled or built-up shapes with noncompact or slender flanges, as determined by AISC Specification Table B4.1b, must be checked for strength based on the limit state of flange local buckling using Equations F6-2 or F6-3 as applicable. The vast majority of W, M, C and MC shapes have compact flanges, and can therefore develop the full plastic moment, Mp, about the minor axis. The provisions of this section are illustrated in Example F.5. F7. SQUARE AND RECTANGULAR HSS AND BOX SECTIONS

Square and rectangular HSS need to be checked for the limit states of yielding, and flange and web local buckling. Lateral-torsional buckling is also possible for rectangular HSS or box sections bent about the strong axis; however, as indicated in the User Note in AISC Specification Section F7, deflection will usually control the design before there is a significant reduction in flexural strength due to lateral-torsional buckling. The design and section property tables in the AISC Manual were calculated using a design wall thickness of 93% of the nominal wall thickness (see AISC Specification Section B4.2). Strength reductions due to local buckling have been accounted for in the AISC Manual design tables. The selection of a square HSS with compact flanges is illustrated in Example F.6. The provisions for a rectangular HSS with noncompact flanges is illustrated in Example F.7. The provisions for a square HSS with slender flanges are illustrated in Example F.8. Available flexural strengths of rectangular and square HSS are listed in Tables 3-12 and 3-13, respectively. If HSS members are specified using ASTM A1065 or ASTM A1085 material, the design wall thickness may be taken equal to the nominal wall thickness. F8. ROUND HSS

The definition of HSS encompasses both tube and pipe products. The lateral-torsional buckling limit state does not apply, but round HSS are subject to strength reductions from local buckling. Available strengths of round HSS and Pipes are listed in AISC Manual Tables 3-14 and 3-15, respectively. The tabulated properties and available flexural strengths of these shapes in the AISC Manual are calculated using a design wall thickness of 93% of the nominal wall thickness. The design of a Pipe is illustrated in Example F.9. If round HSS members are specified using ASTM A1085 material, the design wall thickness may be taken equal to the nominal wall thickness. F9. TEES AND DOUBLE ANGLES LOADED IN THE PLANE OF SYMMETRY

The AISC Specification provides a check for flange local buckling, which applies only when a noncompact or slender flange is in compression due to flexure. This limit state will seldom govern. A check for local buckling of the tee stem in flexural compression was added in the 2010 edition of the Specification. The provisions were expanded to include local buckling of double-angle web legs in flexural compression in the 2016 edition. Attention should be given to end conditions of tees to avoid inadvertent fixed end moments that induce compression in the web unless this limit state is checked. The design of a WT-shape in bending is illustrated in Example F.10. F10. SINGLE ANGLES

Section F10 of the AISC Specification permits the flexural design of single angles using either the principal axes or geometric axes (x- and y-axes). When designing single angles without continuous bracing using the geometric axis design provisions, My must be multiplied by 0.80 for use in Equations F10-1, F10-2 and F10-3. The design of a single angle in bending is illustrated in Example F.11. F11. RECTANGULAR BARS AND ROUNDS

The AISC Manual does not include design tables for these shapes. The local buckling limit state does not apply to any bars. With the exception of rectangular bars bent about the strong axis, solid square, rectangular and round bars are not subject to lateral-torsional buckling and are governed by the yielding limit state only. Rectangular bars bent

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about the strong axis are subject to lateral-torsional buckling and are checked for this limit state with Equations F112 and F11-3, as applicable. These provisions can be used to check plates and webs of tees in connections. A design example of a rectangular bar in bending is illustrated in Example F.12. A design example of a round bar in bending is illustrated in Example F.13. F12. UNSYMMETRICAL SHAPES Due to the wide range of possible unsymmetrical cross sections, specific lateral-torsional and local buckling provisions are not provided in this Specification section. A general template is provided, but appropriate literature investigation and engineering judgment are required for the application of this section. A design example of a Zshaped section in bending is illustrated in Example F.14. F13. PROPORTIONS OF BEAMS AND GIRDERS

This section of the Specification includes a limit state check for tensile rupture due to holes in the tension flange of beams, proportioning limits for I-shaped members, detail requirements for cover plates and connection requirements for built-up beams connected side-to-side. Also included are unbraced length requirements for beams designed using the moment redistribution provisions of AISC Specification Section B3.3.

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EXAMPLE F.1-1A W-SHAPE FLEXURAL MEMBER DESIGN IN MAJOR AXIS BENDING, CONTINUOUSLY BRACED Given: Select a W-shape beam for span and uniform dead and live loads as shown in Figure F.1-1A. Limit the member to a maximum nominal depth of 18 in. Limit the live load deflection to L/360. The beam is simply supported and continuously braced. The beam is ASTM A992 material.

Fig. F.1-1A. Beam loading and bracing diagram. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wu  1.2  0.45 kip/ft   1.6  0.75 kip/ft 

 1.20 kip/ft

 1.74 kip/ft From AISC Manual Table 3-23, Case 1: Mu  

wu L2 8

1.74 kip/ft  35 ft 2

8  266 kip-ft

ASD wa  0.45 kip/ft  0.75 kip/ft

From AISC Manual Table 3-23, Case 1: Ma  

wa L2 8

1.20 kip/ft  35 ft 2

8  184 kip-ft

Required Moment of Inertia for Live-Load Deflection Criterion of L/360  max  

L 360  35 ft 12 in./ft 

360  1.17 in.

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I x ( reqd ) 

5 wL L4 384 E  max

(from AISC Manual Table 3-23, Case 1)

5  0.75 kip/ft  35 ft  12 in./ft  4



3

384  29,000 ksi 1.17 in.

 746 in.4

Beam Selection Select a W1850 from AISC Manual Table 3-3. I x  800 in.4  746 in.4

o.k.

Per the User Note in AISC Specification Section F2, the section is compact. Because the beam is continuously braced and compact, only the yielding limit state applies. From AISC Manual Table 3-2, the available flexural strength is: ASD

LRFD b M n  b M px  379 kip-ft > 266 kip-ft o.k.

M n M px  b b  252 kip-ft > 184 kip-ft

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EXAMPLE F.1-1B W-SHAPE FLEXURAL MEMBER DESIGN IN MAJOR AXIS BENDING, CONTINUOUSLY BRACED Given: Verify the available flexural strength of the ASTM A992 W1850 beam selected in Example F.1-1A by directly applying the requirements of the AISC Specification. Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W1850 Zx = 101 in.3

The required flexural strength from Example F.1-1A is: LRFD M u  266 kip-ft

ASD M a  184 kip-ft

Nominal Flexural Strength Per the User Note in AISC Specification Section F2, the section is compact. Because the beam is continuously braced and compact, only the yielding limit state applies. M n  M p  Fy Z x



(Spec. Eq. F2-1)

  50 ksi  101 in.

3



 5, 050 kip-in. or 421 kip-ft

Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: LRFD b  0.90   b M n  0.90  421 kip-ft 

 379 kip-ft  266 kip-ft o.k.

ASD b  1.67   M n 421 kip-ft  b 1.67  252 kip-ft  184 kip-ft o.k.

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EXAMPLE F.1-2A W-SHAPE FLEXURAL MEMBER DESIGN IN MAJOR AXIS BENDING, BRACED AT THIRD POINTS Given: Use the AISC Manual tables to verify the available flexural strength of the W1850 beam size selected in Example F.1-1A for span and uniform dead and live loads as shown in Figure F.1-2A. The beam is simply supported and braced at the ends and third points. The beam is ASTM A992 material.

Fig. F.1-2A. Beam loading and bracing diagram.

Solution: The required flexural strength at midspan from Example F.1-1A is: LRFD

ASD

M u  266 kip-ft

M a  184 kip-ft

Unbraced Length

35 ft 3  11.7 ft

Lb 

By inspection, the middle segment will govern. From AISC Manual Table 3-1, for a uniformly loaded beam braced at the ends and third points, Cb = 1.01 in the middle segment. Conservatively neglect this small adjustment in this case. Available Flexural Strength Enter AISC Manual Table 3-10 and find the intersection of the curve for the W1850 with an unbraced length of 11.7 ft. Obtain the available strength from the appropriate vertical scale to the left. From AISC Manual Table 3-10, the available flexural strength is: LRFD b M n  302 kip-ft  266 kip-ft

o.k. 

ASD Mn  201 kip-ft  184 kip-ft o.k. b

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EXAMPLE F.1-2B W-SHAPE FLEXURAL MEMBER DESIGN IN MAJOR AXIS BENDING, BRACED AT THIRD POINTS

Given: Verify the available flexural strength of the W1850 beam selected in Example F.1-1A with the beam braced at the ends and third points by directly applying the requirements of the AISC Specification. The beam is ASTM A992 material.

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W1850

ry Sx J rts ho

= 1.65 in. = 88.9 in.3 = 1.24 in.4 = 1.98 in. = 17.4 in.

The required flexural strength from Example F.1-1A is: LRFD

ASD

M u  266 kip-ft

M a  184 kip-ft

Nominal Flexural Strength Calculate Cb. For the lateral-torsional buckling limit state, the nonuniform moment modification factor can be calculated using AISC Specification Equation F1-1. For the center segment of the beam, the required moments for AISC Specification Equation F1-1 can be calculated as a percentage of the maximum midspan moment as: Mmax = 1.00, MA = 0.972, MB = 1.00, and MC = 0.972.

Cb  

12.5M max 2.5M max  3M A  4M B  3M C

(Spec. Eq. F1-1)

12.5 1.00 

2.5 1.00   3  0.972   4 1.00   3  0.972 

 1.01 For the end-span beam segments, the required moments for AISC Specification Equation F1-1 can be calculated as a percentage of the maximum midspan moment as: Mmax = 0.889, MA = 0.306, MB = 0.556, and MC = 0.750. Cb  

2.5M max

12.5M max  3M A  4 M B  3 M C

(Spec. Eq. F1-1)

12.5  0.889 

2.5  0.889   3  0.306   4  0.556   3  0.750 

 1.46

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Thus, the center span, with the higher required strength and lower Cb, will govern. The limiting laterally unbraced length for the limit state of yielding is:

L p  1.76ry

E Fy

(Spec. Eq. F2-5)

29, 000 ksi 50 ksi  69.9 in. or 5.83 ft  1.76 1.65 in.

The limiting unbraced length for the limit state of inelastic lateral-torsional buckling, with c = 1 from AISC Specification Equation F2-8a for doubly symmetric I-shaped members, is:

Lr  1.95rts

E 0.7 Fy

2

Jc  0.7 Fy   Jc       6.76  S x ho  S x ho   E 

 29, 000 ksi   1.95 1.98 in.    0.7  50 ksi  

2

(Spec. Eq. F2-6)

1.24 in.  1.0    1.24 in.  1.0   88.9 in.  17.4 in.  88.9 in.  17.4 in.  4

4

3

3

2

 0.7  50 ksi    6.76    29, 000 ksi 

2

 203 in. or 16.9 ft

For a compact beam with an unbraced length of Lp  Lb  Lr, the lesser of either the flexural yielding limit state or the inelastic lateral-torsional buckling limit state controls the nominal strength. Mp = 5,050 kip-in. (from Example F.1-1B)   Lb  L p   M n  Cb  M p  ( M p  0.7 Fy S x )  (Spec. Eq. F2-2)    M p   Lr  L p     11.7 ft  5.83 ft    1.01 5, 050 kip-in.  5, 050 kip-in.  0.7  50 ksi  88.9 in.3      5, 050 kip-in.    16.9 ft  5.83 ft     4, 060 kip-in.  5, 050 kip-in.  4, 060 kip-in. or 339 kip-ft





Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: LRFD b  0.90   b M n  0.90  339 kip-ft 

 305 kip-ft  266 kip-ft o.k.



ASD b  1.67   M n 339 kip-ft  b 1.67  203 kip-ft  184 kip-ft o.k.

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EXAMPLE F.1-3A W-SHAPE FLEXURAL MEMBER DESIGN IN MAJOR AXIS BENDING, BRACED AT MIDSPAN Given: Use the AISC Manual tables to verify the available flexural strength of the W1850 beam size selected in Example F.1-1A for span and uniform dead and live loads as shown in Figure F.1-3A. The beam is simply supported and braced at the ends and midpoint. The beam is ASTM A992 material.

Fig. F.1-3A. Beam loading and bracing diagram.

Solution:

The required flexural strength at midspan from Example F.1-1A is: LRFD

ASD

M u  266 kip-ft

M a  184 kip-ft

Unbraced Length

35 ft 2  17.5 ft

Lb 

From AISC Manual Table 3-1, for a uniformly loaded beam braced at the ends and at the center point, Cb = 1.30. There are several ways to make adjustments to AISC Manual Table 3-10 to account for Cb greater than 1.0. Procedure A Available moments from the sloped and curved portions of the plots from AISC Manual Table 3-10 may be multiplied by Cb, but may not exceed the value of the horizontal portion (Mp for LRFD, Mp/ for ASD). Obtain the available strength of a W1850 with an unbraced length of 17.5 ft from AISC Manual Table 3-10. Enter AISC Manual Table 3-10 and find the intersection of the curve for the W1850 with an unbraced length of 17.5 ft. Obtain the available strength from the appropriate vertical scale to the left. LRFD

ASD

b M n  222 kip-ft 

Mn  148 kip-ft  b

From AISC Manual Table 3-2:

From AISC Manual Table 3-2:

b M p  379 kip-ft (upper limit on Cb b M n ) 

Mp b

 252 kip-ft (upper limit on Cb

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LRFD

ASD

Adjust for Cb.

Adjust for Cb.

1.30  222 kip-ft   289 kip-ft

1.30 148 kip-ft   192 kip-ft

Check limit.

Check limit.

289 kip-ft  b M p  379 kip-ft

o.k.

192 kip-ft 

Mp b

 252 kip-ft o.k.

Check available versus required strength.

Check available versus required strength.

289 kip-ft  266 kip-ft o.k.

192 kip-ft  184 kip-ft o.k.

Procedure B For preliminary selection, the required strength can be divided by Cb and directly compared to the strengths in AISC Manual Table 3-10. Members selected in this way must be checked to ensure that the required strength does not exceed the available plastic moment strength of the section. Calculate the adjusted required strength. LRFD

ASD

266 kip-ft M u  1.30  205 kip-ft

184 kip-ft M a  1.30  142 kip-ft

Obtain the available strength for a W1850 with an unbraced length of 17.5 ft from AISC Manual Table 3-10. LRFD

ASD

Mn  148 kip-ft  142 kip-ft o.k.  b

b M n  222 kip-ft  205 kip-ft o.k.  b M p  379 kip-ft  266 kip-ft

 o.k.

Mp b

 252 kip-ft  184 kip-ft o.k.

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EXAMPLE F.1-3B W-SHAPE FLEXURAL MEMBER DESIGN IN MAJOR-AXIS BENDING, BRACED AT MIDSPAN Given:

Verify the available flexural strength of the W1850 beam selected in Example F.1-1A with the beam braced at the ends and center point by directly applying the requirements of the AISC Specification. The beam is ASTM A992 material. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W1850

rts Sx J ho

= 1.98 in. = 88.9 in.3 = 1.24 in.4 = 17.4 in.

The required flexural strength from Example F.1-1A is: LRFD

ASD

M u  266 kip-ft

M a  184 kip-ft

Nominal Flexural Strength Calculate Cb. The required moments for AISC Specification Equation F1-1 can be calculated as a percentage of the maximum midspan moment as: Mmax = 1.00, MA = 0.438, MB = 0.750, and MC = 0.938.

Cb  

12.5M max 2.5M max  3M A  4M B  3M C

(Spec. Eq. F1-1)

12.5 1.00 

2.5 1.00   3  0.438  4  0.750   3  0.938

 1.30 From AISC Manual Table 3-2: Lp = 5.83 ft Lr = 16.9 ft From Example F.1-3A: Lb = 17.5 ft

For a compact beam with an unbraced length Lb > Lr, the limit state of elastic lateral-torsional buckling applies.

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Calculate Fcr, where c = 1.0 for doubly symmetric I-shapes. Fcr 



Cb 2 E  Lb  r   ts 

2

1  0.078

Jc  Lb  S x ho  rts 

1.302  29, 000 ksi 

 (17.5ft)(12 in./ft)    1.98 in.  43.2 ksi

2

2

(Spec. Eq. F2-4)

1.24 in.  1.0   17.5 ft 12 in./ft     88.9 in.  17.4 in.  1.98 in.  4

1  0.078

2

3

M p  5,050 kip-in. (from Example F.1-1B) M n  Fcr S x  M p



(Spec. Eq. F2-3)



  43.2 ksi  88.9 in.3  5,050 kip-in.  3,840 kip-in.  5,050 kip-in.  3,840 kip-in. or 320 kip-ft Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: LRFD

ASD

b  0.90   b M n  0.90  320 kip-ft 

 288 kip-ft  266 kip-ft o.k.



b  1.67   M n 320 kip-ft   b 1.67  192 kip-ft  184 kip-ft o.k.

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EXAMPLE F.2-1A COMPACT CHANNEL FLEXURAL MEMBER, CONTINUOUSLY BRACED Given: Using the AISC Manual tables, select a channel to serve as a roof edge beam for span and uniform dead and live loads as shown in Figure F.2-1A. The beam is simply supported and continuously braced. Limit the live load deflection to L/360. The channel is ASTM A36 material.

Fig. F.2-1A. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wu  1.2  0.23 kip/ft   1.6  0.69 kip/ft 

 0.920 kip/ft

 1.38 kip/ft From AISC Manual Table 3-23, Case 1: Mu  

ASD wa  0.23 kip/ft  0.69 kip/ft

wu L2 8

From AISC Manual Table 3-23, Case 1: Ma 

1.38 kip/ft  25 ft 2



wa L2 8

 0.920 kip/ft  25 ft 2

8  71.9 kip-ft

8  108 kip-ft

Beam Selection Per the User Note in AISC Specification Section F2, all ASTM A36 channels are compact. Because the beam is compact and continuously braced, the yielding limit state governs and Mn = Mp. Try C1533.9 from AISC Manual Table 3-8. LRFD

b M n  b M p  137 kip-ft  108 kip-ft o.k.

ASD Mn M p  b b  91.3 kip-ft  71.9 kip-ft o.k.

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Live Load Deflection Limit the live load deflection at the center of the beam to L/360.  max  

L 360  25 ft 12 in./ft 

360  0.833 in.

For C1533.9, Ix = 315 in.4 from AISC Manual Table 1-5. The maximum calculated deflection is:   max 

5wL L4 384EI

(from AISC Manual Table 3-23, Case 1)

5  0.69 kip/ft  25 ft  12 in./ft  4





384  29,000 ksi  315 in.4

3



 0.664 in.  0.833 in. o.k.

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EXAMPLE F.2-1B COMPACT CHANNEL FLEXURAL MEMBER, CONTINUOUSLY BRACED Given:

Verify the available flexural strength of the C1533.9 beam selected in Example F.2-1A by directly applying the requirements of the AISC Specification. The channel is ASTM A36 material. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-5, the geometric properties are as follows: C1533.9

Zx = 50.8 in.3 The required flexural strength from Example F.2-1A is: LRFD

ASD

M u  108 kip-ft

M a  71.9 kip-ft

Nominal Flexural Strength Per the User Note in AISC Specification Section F2, all ASTM A36 C- and MC-shapes are compact. A channel that is continuously braced and compact is governed by the yielding limit state. M n  M p  Fy Z x



(Spec. Eq. F2-1)

  36 ksi  50.8 in.

3



 1,830 kip-in. or 152 kip-ft

Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: LRFD b  0.90   b M n  0.90 152 kip-ft 

 137 kip-ft  108 kip-ft o.k.



ASD b  1.67   M n 152 kip-ft   b 1.67  91.0 kip-ft  71.9 kip-ft o.k.

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EXAMPLE F.2-2A COMPACT CHANNEL FLEXURAL MEMBER WITH BRACING AT ENDS AND FIFTH POINTS Given: Use the AISC Manual tables to verify the available flexural strength of the C1533.9 beam selected in Example F.2-1A for span and uniform dead and live loads as shown in Figure F.2-2A. The beam is simply supported and braced at the ends and fifth points. The channel is ASTM A36 material.

Fig. F.2-2A. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi The center segment will govern by inspection. The required flexural strength at midspan from Example F.2-1A is: LRFD

ASD

M u  108 kip-ft

M a  71.9 kip-ft

From AISC Manual Table 3-1, with an almost uniform moment across the center segment, Cb = 1.00; therefore, no adjustment is required. Unbraced Length

25ft 5  5.00 ft

Lb 

Obtain the strength of the C1533.9 with an unbraced length of 5.00 ft from AISC Manual Table 3-11. Enter AISC Manual Table 3-11 and find the intersection of the curve for the C1533.9 with an unbraced length of 5.00 ft. Obtain the available strength from the appropriate vertical scale to the left. LRFD

ASD

b M n  130 kip-ft  108 kip-ft o.k. 

Mn  87.0 kip-ft  71.9 kip-ft o.k. b

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EXAMPLE F.2-2B COMPACT CHANNEL FLEXURAL MEMBER WITH BRACING AT ENDS AND FIFTH POINTS Given:

Verify the results from Example F.2-2A by directly applying the requirements of the AISC Specification. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-5, the geometric properties are as follows: C1533.9

Sx = 42.0 in.3

The required flexural strength from Example F.2-1A is: LRFD

ASD

M u  108 kip-ft

M a  71.9 kip-ft

Available Flexural Strength Per the User Note in AISC Specification Section F2, all ASTM A36 C- and MC-shapes are compact. From AISC Manual Table 3-1, for the center segment of a uniformly loaded beam braced at the ends and the fifth points: Cb = 1.00 From AISC Manual Table 3-8, for a C1533.9: Lp = 3.75 ft Lr = 14.5 ft From Example F2.2A: Lb = 5.00 ft For a compact channel with Lp < Lb ≤ Lr, the lesser of the flexural yielding limit state or the inelastic lateral-torsional buckling limit state controls the available flexural strength. The nominal flexural strength based on the flexural yielding limit state, from Example F.2-1B, is:

Mn  M p  1,830 kip-in.

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The nominal flexural strength based on the lateral-torsional buckling limit state is:   Lb  L p   M n  Cb  M p   M p  0.7 Fy S x   (Spec. Eq. F2-2)    M p  Lr  L p      5.00 ft  3.75 ft    1.00 1,830 kip-in.  1,830 kip-in.  0.7  36 ksi  42.0 in.3      1,830 kip-in.    14.5 ft  3.75 ft    =1,740 kip-in.  1,830 kip-in. =1,740 kip-in. or 145 kip-ft





Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: LRFD b  0.90   b M n  0.90 145 kip-ft 

 131 kip-ft  108 kip-ft o.k.



ASD b  1.67   M n 145 kip-ft   b 1.67  86.8 kip-ft  71.9 kip-ft o.k.

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EXAMPLE F.3A W-SHAPE FLEXURAL MEMBER WITH NONCOMPACT FLANGES IN MAJOR AXIS BENDING Given: Using the AISC Manual tables, select a W-shape beam for span, uniform dead load, and concentrated live loads as shown in Figure F.3A. The beam is simply supported and continuously braced. Also calculate the deflection. The beam is ASTM A992 material.

Fig. F.3A. Beam loading and bracing diagram.

Note: A beam with noncompact flanges will be selected to demonstrate that the tabulated values of the AISC Manual account for flange compactness. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength at midspan is:

wu  1.2  0.05 kip/ft 

LRFD

ASD

wa  0.05 kip/ft

 0.0600 kip/ft

Pu  1.6 18 kips 

Pa  18 kips

 28.8 kips From AISC Manual Table 3-23, Cases 1 and 9: Mu  

wu L2  Pu a 8

 0.0600 kip/ft  40 ft 2

 396 kip-ft

8

From AISC Manual Table 3-23, Cases 1 and 9: Ma 

 40 ft    28.8 kips     3 



wa L2  Pa a 8

 0.05 kip/ft  40 ft 2

8  250 kip-ft

 40 ft   18 kips     3 

Beam Selection For a continuously braced W-shape, the available flexural strength equals the available plastic flexural strength.

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Select the lightest section providing the required strength from the bold entries in AISC Manual Table 3-2. Try a W2148. This beam has a noncompact compression flange at Fy = 50 ksi as indicated by footnote “f” in AISC Manual Table 3-2. This shape is also footnoted in AISC Manual Table 1-1. From AISC Manual Table 3-2, the available flexural strength is: LRFD

ASD M Mn px  b b  265 kip-ft > 250 kip-ft o.k.

b M n  b M px  398 kip-ft > 396 kip-ft o.k.

Note: The value Mpx in AISC Manual Table 3-2 includes the strength reductions due to the shape being noncompact. Deflection From AISC Manual Table 1-1: Ix = 959 in.4 The maximum deflection occurs at the center of the beam.  max 

5wD L4 23PL L3  384EI 648EI

(AISC Manual Table 3-23, Cases 1 and 9)

5  0.05 kip/ft  40 ft  12 in./ft  4





384  29,000 ksi  959 in.4

3



23 18 kips  40 ft  12 in./ft  3





648  29,000 ksi  959 in.4

3



 2.64 in.

This deflection can be compared with the appropriate deflection limit for the application. Deflection will often be more critical than strength in beam design.

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EXAMPLE F.3B W-SHAPE FLEXURAL MEMBER WITH NONCOMPACT FLANGES IN MAJOR AXIS BENDING Given:

Verify the results from Example F.3A by directly applying the requirements of the AISC Specification. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W2148

Sx = 93.0 in.3 Zx = 107 in.3 bf = 9.47 2t f

The required flexural strength from Example F.3A is: ASD

LRFD M u  396 kip-ft

M a  250 kip-ft

Flange Slenderness  bf   2t f  9.47

The limiting width-to-thickness ratios for the compression flange are:   pf  0.38  0.38

E Fy

(Spec. Table B4.1b, Case 10)

29,000 ksi 50 ksi

 9.15

  rf  1.0  1.0

E Fy

(Spec. Table B4.1b, Case 10)

29,000 ksi 50 ksi

 24.1

pf <  < rf, therefore, the compression flange is noncompact. This could also be determined from the footnote “f” in AISC Manual Table 1-1.

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Nominal Flexural Strength Because the beam is continuously braced, and therefore not subject to lateral-torsional buckling, the available strength is based on the limit state of compression flange local buckling. From AISC Specification Section F3.2: M p  Fy Z x

(Spec. Eq. F2-1)



  50 ksi  107 in.3



 5,350 kip-in. or 446 kip-ft

     pf   M n   M p   M p  0.7 Fy S x        rf   pf     9.47  9.15    5,350 kip-in.  5,350 kip-in.  0.7  50 ksi  93.0 in.3       24.1  9.15     5,310 kip-in. or 442 kip-ft



(Spec. Eq. F3-1)



Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: LRFD b  0.90   b M n  0.90  442 kip-ft 

 398 kip-ft  396 kip-ft o.k.

ASD

b  1.67 M n 442 kip-ft  1.67 b  265 kip-ft  250 kip-ft o.k.

Note that these available strengths are identical to the tabulated values in AISC Manual Table 3-2, as shown in Example F.3A, which account for the noncompact flange.

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EXAMPLE F.4 W-SHAPE FLEXURAL MEMBER, SELECTION BY MOMENT OF INERTIA FOR MAJOR AXIS BENDING Given: Using the AISC Manual tables, select a W-shape using the moment of inertia required to limit the live load deflection to 1.00 in. for span and uniform dead and live loads as shown in Figure F.4. The beam is simply supported and continuously braced. The beam is ASTM A992 material.

Fig. F.4. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wu  1.2  0.8 kip/ft   1.6  2 kip/ft 

ASD

wa  0.8 kip/ft  2 kip/ft  2.80 kip/ft

 4.16 kip/ft From AISC Manual Table 3-23, Case 1: Mu  

wu L2 8

From AISC Manual Table 3-23, Case 1: Ma 

 4.16 kip/ft  30 ft 2



wa L2 8

 2.80 kip/ft  30 ft 2

8  315 kip-ft

8  468 kip-ft

Minimum Required Moment of Inertia The maximum live load deflection, max, occurs at midspan and is calculated as:  max 

5wL L4 384EI

(AISC Manual Table 3-23, Case 1)

Rearranging and substituting max = 1.00 in.,

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I min 

5wL L4 384 E  max 5  2 kip/ft  30 ft  12 in./ft  4



3

384  29, 000 ksi 1.00 in.

 1, 260 in.4 Beam Selection Select the lightest section with the required moment of inertia from the bold entries in AISC Manual Table 3-3. Try a W2455. Ix = 1,350 in.4 > 1,260 in.4

o.k.

Because the W2455 is continuously braced and compact, its strength is governed by the yielding limit state and AISC Specification Section F2.1. From AISC Manual Table 3-2, the available flexural strength is: LRFD b M n  b M px  503 kip-ft > 468 kip-ft o.k.

ASD M n M px  b b  334 kip-ft > 315 kip-ft o.k.

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EXAMPLE F.5

I-SHAPED FLEXURAL MEMBER IN MINOR AXIS BENDING

Given:

Using the AISC Manual tables, select a W-shape beam loaded on its minor axis for span and uniform dead and live loads as shown in Figure F.5. Limit the live load deflection to L/240. The beam is simply supported and braced only at the ends. The beam is ASTM A992 material.

Fig. F.5. Beam loading and bracing diagram.

Note: Although not a common design case, this example is being used to illustrate AISC Specification Section F6 (Ishaped members and channels bent about their minor axis). Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wu  1.2  0.667 kip/ft   1.6  2 kip/ft 

ASD wa  0.667 kip/ft  2 kip/ft

 2.67 kip/ft

 4.00 kip/ft From AISC Manual Table 3-23, Case 1: Mu  

wu L2 8

From AISC Manual Table 3-23, Case 1: Ma 

 4.00 kip/ft 15 ft 2



wa L2 8

 2.67 kip/ft 15 ft 2

8  75.1 kip-ft

8

 113 kip-ft

Minimum Required Moment of Inertia The maximum live load deflection permitted is:  max 





L 240 15 ft 12 in./ft 

240  0.750 in.

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 I y , reqd 

5 wL L4 384 E  max

(modified AISC Manual Table 3-23, Case 1)

5  2 kip/ft 15 ft  12 in./ft  4



3

384  29, 000 ksi  0.750 in.

 105 in.4

Beam Selection Select the lightest section from the bold entries in AISC Manual Table 3-5. Try a W1258. From AISC Manual Table 1-1, the geometric properties are as follows: W1258

Sy = 21.4 in.3 Zy = 32.5 in.3 Iy = 107 in.4 > 105 in.4 o.k. (for deflection requirement) Nominal Flexural Strength AISC Specification Section F6 applies. Because the W1258 has compact flanges per the User Note in this Section, the yielding limit state governs the design.

M n  M p  Fy Z y  1.6 Fy S y







  50 ksi  32.5 in.3  1.6  50 ksi  21.4 in.3

(Spec. Eq. F6-1)



 1, 630 kip-in.  1,710 kip-in.  1, 630 kip-in or 136 kip-ft Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: LRFD

ASD

b  0.90   b M n  0.90 136 kip-ft 

 122 kip-ft  113 kip-ft o.k. 

b  1.67 

M n 136 kip-ft  1.67 b   81.4 kip-ft  75.1 kip-ft o.k.

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EXAMPLE F.6

SQUARE HSS FLEXURAL MEMBER WITH COMPACT FLANGES

Given: Using the AISC Manual tables, select a square HSS beam for span and uniform dead and live loads as shown in Figure F.6. Limit the live load deflection to L/240. The beam is simply supported and continuously braced. The HSS is ASTM A500 Grade C material.

Fig. F.6. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C, rectangular HSS Fy = 50 ksi Fu = 62 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wu  1.2  0.145 kip/ft   1.6  0.435 kip/ft 

ASD wa  0.145 kip/ft  0.435 kip/ft

 0.580 kip/ft

 0.870 kip/ft From AISC Manual Table 3-23, Case 1: Mu  

wu L2 8

From AISC Manual Table 3-23, Case 1: Ma 

 0.870 kip/ft  7.5 ft 2



8

wa L2 8

 0.580 kip/ft  7.5 ft 2 8

 4.08 kip-ft

 6.12 kip-ft

Minimum Required Moment of Inertia The maximum live load deflection permitted is:  max  

L 240  7.5 ft 12 in./ft 

240  0.375 in.

Determine the minimum required moment of inertia as follows.

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I req 

5 wL L4 384 E  max

(from AISC Manual Table 3-23, Case 1)

5  0.435 kip/ft  7.5 ft  12 in./ft  4



3

384  29, 000 ksi  0.375 in.

 2.85 in.4

Beam Selection Select an HSS with a minimum Ix of 2.85 in.4, using AISC Manual Table 1-12, and having adequate available strength, using AISC Manual Table 3-13. Try an HSS32328. From AISC Manual Table 1-12, I x  2.90 in.4  2.85 in.4

o.k.

From AISC Manual Table 3-13, the available flexural strength is: ASD

LRFD b M n  7.21 kip-ft > 6.12 kip-ft o.k.

Mn  4.79 kip-ft  4.08 kip-ft o.k. b

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EXAMPLE F.7A RECTANGULAR HSS FLEXURAL MEMBER WITH NONCOMPACT FLANGES Given:

Using the AISC Manual tables, select a rectangular HSS beam for span and uniform dead and live loads as shown in Figure F.7A. Limit the live load deflection to L/240. The beam is simply supported and braced at the end points only. A noncompact member was selected here to illustrate the relative ease of selecting noncompact shapes from the AISC Manual, as compared to designing a similar shape by applying the AISC Specification requirements directly, as shown in Example F.7B. The HSS is ASTM A500 Grade C material.

Fig. F.7A. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C, rectangular HSS Fy = 50 ksi Fu = 62 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wu  1.2  0.15 kip/ft   1.6  0.4 kip/ft 

 0.820 kip/ft From AISC Manual Table 3-23, Case 1: Mu  

wu L2 8

ASD wa  0.15 kip/ft  0.4 kip/ft  0.550 kip/ft From AISC Manual Table 3-23, Case 1: Ma 

 0.820 kip/ft  21 ft 2



wa L2 8

 0.550 kip/ft  21 ft 2

8  30.3 kip-ft

8  45.2 kip-ft

Minimum Required Moment of Inertia The maximum live load deflection permitted is:  max  

L 240  21 ft 12 in./ft 

240  1.05 in.

Determine the minimum required moment of inertia as follows:

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

I min 

5wL L4 384 E  max

(from AISC Manual Table 3-23, Case 1)

5  0.4 kip/ft  21 ft  12 in./ft  4



3

384  29, 000 ksi 1.05 in.

 57.5 in.4 Beam Selection Select a rectangular HSS with a minimum Ix of 57.5 in.4, using AISC Manual Table 1-11, and having adequate available strength, using AISC Manual Table 3-12. Try an HSS106x oriented in the strong direction. This rectangular HSS section was purposely selected for illustration purposes because it has a noncompact flange. See AISC Manual Table 1-12A for compactness criteria. I x  74.6 in.4  57.5 in.4

o.k.

From AISC Manual Table 3-12, the available flexural strength is: LRFD b M n  59.7 kip-ft > 45.2 kip-ft o.k.

ASD Mn  39.7 kip-ft  30.3 kip-ft o.k. b

Note: Because AISC Manual Table 3-12 does not account for lateral-torsional buckling, it needs to be checked using AISC Specification Section F7.4. As discussed in the User Note to AISC Specification Section F7.4, lateral-torsional buckling will not occur in square sections or sections bending about their minor axis. In HSS sizes, deflection will often occur before there is a significant reduction in flexural strength due to lateral-torsional buckling. See Example F.7B for the calculation accounting for lateral-torsional buckling for the HSS106x.

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EXAMPLE F.7B RECTANGULAR HSS FLEXURAL MEMBER WITH NONCOMPACT FLANGES Given:

In Example F.7A the required information was easily determined by consulting the tables of the AISC Manual. The purpose of the following calculation is to demonstrate the use of the AISC Specification to calculate the flexural strength of an HSS member with a noncompact compression flange. The HSS is ASTM A500 Grade C material. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C, rectangular HSS Fy = 50 ksi Fu = 62 ksi From AISC Manual Table 1-11, the geometric properties are as follows: HSS106x

= 5.37 in.2 = 18.0 in.3 = 14.9 in.3 = 2.52 in. = 73.8 in.4 = 31.5 = 54.5

Ag Zx Sx ry J b/t h/t

Flange Compactness 

b tf

b t  31.5 

From AISC Specification Table B4.1b, Case 17, the limiting width-to-thickness ratios for the flange are:   p  1.12  1.12

E Fy 29, 000 ksi 50 ksi

 27.0  r  1.40



 1.40

E Fy 29, 000 ksi 50 ksi

 33.7

p <  < r; therefore, the flange is noncompact and AISC Specification Equation F7-2 applies.

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Web Compactness 

h t  54.5



From AISC Specification Table B4.1b, Case 19, the limiting width-to-thickness ratio for the web is:  p  2.42



 2.42

E Fy 29, 000 ksi 50 ksi

 58.3    p ; therefore, the web is compact and the limit state of web local buckling does not apply.

Nominal Flexural Strength Flange Local Buckling From AISC Specification Section F7.2(b), the limit state of flange local buckling applies for HSS with noncompact flanges and compact webs. M p  Fy Z x



  50 ksi  18.0 in.3

 from Spec. Eq. F7-1



 900 kip-in.

 b M n  M p   M p  Fy S   3.57  t f 

 Fy  4.0   M p  E 

(Spec. Eq. F7-2)

  50 ksi  900 kip-in.  900 kip-in.   50 ksi  14.9 in.3  3.57  31.5  4.0  900 kip-in.   29, 000 ksi    796 kip-in.  900 kip-in.





 796 kip-in. or 66.4 kip-ft Yielding and Lateral-Torsional Buckling Determine the limiting laterally unbraced lengths for the limit state of yielding and the limit state of inelastic lateraltorsional buckling using AISC Specification Section F7.4.

Lb   21 ft 12 in./ft   252 in. L p  0.13Ery

JAg

(Spec. Eq. F7-12)

Mp

 73.8 in. 5.37 in.  4

 0.13  29, 000 ksi  2.52 in.

2

900 kip-in.

 210 in.

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Lr  2 Ery

JAg

(Spec. Eq. F7-13)

0.7 Fy S x

 73.8 in.  5.37 in.  0.7  50 ksi  14.9 in.  4

 2  29, 000 ksi  2.52 in.

2

3

 5, 580 in.

For the lateral-torsional buckling limit state, the lateral-torsional buckling modification factor can be calculated using AISC Specification Equation F1-1. For the beam, the required moments for AISC Specification Equation F1-1 can be calculated as a percentage of the maximum midspan moment as: Mmax = 1.00, MA = 0.750, MB = 1.00, and MC = 0.750.

Cb  

12.5M max 2.5M max  3M A  4M B  3M C

(Spec. Eq. F1-1)

12.5 1.00 

2.5 1.00   3  0.750   4 1.00   3  0.750 

 1.14 Since L p  Lb  Lr , the nominal moment strength considering lateral-torsional buckling is given by:   Lb  L p M n  Cb  M p   M p  0.7 Fy S x    Lr  L p 

    M p  

(Spec. Eq. F7-10)

  252 in.  210 in.    1.14 900 kip-in.  900 kip-in.  0.7  50 ksi  14.9 in.3      900 kip-in.    5,580 in.  210 in.     1, 020 kip-in.  900 kip-in.  900 kip-in. or 75.0 kip-ft





Available Flexural Strength The nominal strength is controlled by flange local buckling and therefore: M n  66.4 kip-ft From AISC Specification Section F1, the available flexural strength is: ASD

LRFD b  0.90   b M n  0.90  66.4 kip-ft 

 59.8 kip-ft

b  1.67 

M n 66.4 kip-ft  1.67 b   39.8 kip-ft

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EXAMPLE F.8A SQUARE HSS FLEXURAL MEMBER WITH SLENDER FLANGES Given:

Using AISC Manual tables, verify the strength of an HSS88x beam for span and uniform dead and live loads as shown in Figure F.8A. Limit the live load deflection to L/240. The beam is simply supported and continuously braced. The HSS is ASTM A500 Grade C material.

Fig. F.8A. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi From AISC Manual Table 1-12, the geometric properties are as follows: HSS88x

Ix = Iy = 54.4 in.4

From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wu  1.2  0.125 kip/ft   1.6  0.375 kip/ft 

 0.750 kip/ft From AISC Manual Table 3-23, Case 1: Mu  

wu L2 8

ASD wa  0.125 kip/ft  0.375 kip/ft  0.500 kip/ft From AISC Manual Table 3-23, Case 1: Ma 

 0.750 kip/ft  21.0 ft 2



8

wa L2 8

 0.500 kip/ft  21.0 ft 2 8

 27.6 kip-ft

 41.3 kip-ft

From AISC Manual Table 3-13, the available flexural strength is: LRFD

ASD

b M n  46.3 kip-ft > 41.3 kip-ft o.k.

Mn  30.8 kip-ft  27.6 kip-ft o.k. b

Note that the strengths given in AISC Manual Table 3-13 incorporate the effects of noncompact and slender elements.

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Deflection The maximum live load deflection permitted is:  max  

L 240  21.0 ft 12 in./ft  240

 1.05 in.

The calculated deflection is: 

5wL L4 384 EI

(modified AISC Manual Table 3-23 Case 1)

5  0.375 kip/ft  21.0 ft  12 in./ft  4





384  29, 000 ksi  54.4 in.4

3



 1.04 in.  1.05 in. o.k.

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EXAMPLE F.8B SQUARE HSS FLEXURAL MEMBER WITH SLENDER FLANGES Given:

In Example F.8A the available strengths were easily determined from the tables of the AISC Manual. The purpose of the following calculation is to demonstrate the use of the AISC Specification to calculate the flexural strength of the HSS beam given in Example F.8A. The HSS is ASTM A500 Grade C material. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C, rectangular HSS Fy = 50 ksi Fu = 62 ksi From AISC Manual Table 1-12, the geometric properties are as follows: HSS88x

I = 54.4 in.4 Z = 15.7 in.3 S = 13.6 in.3 B = 8.00 in. H = 8.00 in. t = 0.174 in. b/t = 43.0 h/t = 43.0 The required flexural strength from Example F.8A is: LRFD

ASD

M u  41.3 kip-ft

M a  27.6 kip-ft

Flange Slenderness The outside corner radii of HSS shapes are taken as 1.5t and the design thickness is used in accordance with AISC Specification Section B4.1b to check compactness. Determine the limiting ratio for a slender HSS flange in flexure from AISC Specification Table B4.1b, Case 17.  r  1.40  1.40

E Fy 29, 000 ksi  50 ksi

 33.7 b t b  tf



 43.0   r ; therefore, the flange is slender

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Web Slenderness Determine the limiting ratio for a compact web in flexure from AISC Specification Table B4.1b, Case 19.  p  2.42  2.42

E Fy 29, 000 ksi 50 ksi

 58.3 h t  43.0   p ; therefore, the web is compact and the limit state of web local buckling does not apply



Nominal Flexural Strength Flange Local Buckling For HSS sections with slender flanges and compact webs, AISC Specification Section F7.2(c) applies. M n  Fy S e

(Spec. Eq. F7-3)

From AISC Specification Section B4.1b(d), the width of the compression flange is determined as follows:

b  8.00 in.  3  0.174 in.  7.48 in. Where the effective section modulus, Se, is determined using the effective width of the compression flange as follows:

be  1.92t f

E Fy

 0.38 1  /tf b 

 1.92  0.174 in.

   b  29, 000 ksi   0.38  29, 000 ksi  1     7.48 in.  50 ksi   43.0  50 ksi 

E Fy

(Spec. Eq. F7-4)

 6.33 in. The ineffective width of the compression flange is:

b  be  7.48 in.  6.33 in.  1.15 in. An exact calculation of the effective moment of inertia and section modulus could be performed taking into account the ineffective width of the compression flange and the resulting neutral axis shift. Alternatively, a simpler but slightly conservative calculation can be performed by removing the ineffective width symmetrically from both the top and bottom flanges.

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 bt 3  I eff  I x      ad 2  12   2  1.15 in. 0.174 in.3  8.00 in.  0.174 in.    54.4 in.4  2   1.15 in. 0.174 in.    12 2    

 48.3 in.4

The effective section modulus is calculated as follows: Se 



I eff H    2 48.3 in.4  8.00 in.     2 

 12.1 in.3

M n  Fy Se

(Spec. Eq. F7-3)



  50 ksi  12.1 in.

3



 605 kip-in. or 50.4 kip-ft

Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: LRFD

ASD

b  0.90

b  1.67

b M n  0.90  50.4 kip-ft 

M n 50.4 kip-ft  1.67 b  30.2 kip-ft  27.6 kip-ft o.k.

 45.4 kip-ft  41.3 kip-ft o.k.

Note that the calculated available strengths are somewhat lower than those in AISC Manual Table 3-13 due to the use of the conservative calculation of the effective section modulus. Also, note that per the User Note in AISC Specification Section F7.4, lateral-torsional buckling is not applicable to square HSS.

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EXAMPLE F.9A PIPE FLEXURAL MEMBER Given:

Using AISC Manual tables, select a Pipe shape with an 8-in. nominal depth for span and uniform dead and live loads as shown in Figure F.9A. There is no deflection limit for this beam. The beam is simply supported and braced at end points only. The Pipe is ASTM A53 Grade B material.

Fig. F.9A. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A53 Grade B Fy = 35 ksi Fu = 60 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wu  1.2  0.32 kip/ft   1.6  0.96 kip/ft 

 1.92 kip/ft From AISC Manual Table 3-23, Case 1: Mu  

wu L2 8

ASD wa  0.32 kip/ft  0.96 kip/ft  1.28 kip/ft From AISC Manual Table 3-23, Case 1: Ma 

1.92 kip/ft 16 ft 2



wa L2 8

1.28 kip/ft 16 ft 2

8  41.0 kip-ft

8  61.4 kip-ft

Pipe Selection Select a member from AISC Manual Table 3-15 having the required strength. Select Pipe 8 x-Strong. From AISC Manual Table 3-15, the available flexural strength is: LRFD b M n  81.4 kip-ft > 61.4 kip-ft o.k.

ASD Mn  54.1 kip-ft  41.0 kip-ft o.k. b

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EXAMPLE F.9B PIPE FLEXURAL MEMBER Given:

The available strength in Example F.9A was easily determined using AISC Manual Table 3-15. The following example demonstrates the calculation of the available strength by directly applying the AISC Specification. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A53 Grade B Fy = 35 ksi Fu = 60 ksi From AISC Manual Table 1-14, the geometric properties are as follows: Pipe 8 x-Strong

Z = 31.0 in.3 D/t = 18.5

The required flexural strength from Example F.9A is: LRFD M u  61.4 kip-ft

ASD M a  41.0 kip-ft

Slenderness Check Determine the limiting diameter-to-thickness ratio for a compact section from AISC Specification Table B4.1b Case 20.  p  0.07

E Fy

 29, 000 ksi   0.07    35 ksi   58.0 D t  18.5   p ; therefore, the section is compact and the limit state of flange local buckling does not apply



0.45E 0.45  29, 000 ksi   35 ksi Fy  373  18.5; therefore, AISC Specification Section F8 applies

Nominal Flexural Strength Based on the limit state of yielding given in AISC Specification Section F8.1:

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M n  M p  Fy Z



(Spec. Eq. F8-1)

  35 ksi  31.0 in.3



 1, 090 kip-in. or 90.4 kip-ft

Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: LRFD

ASD

b  0.90

b  1.67

b M n  0.90  90.4 kip-ft 

M n 90.4 kip-ft  1.67 b  54.1 kip-ft  41.0 kip-ft o.k.

 81.4 kip-ft  61.4 kip-ft o.k.

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EXAMPLE F.10 WT-SHAPE FLEXURAL MEMBER Given:

Directly applying the requirements of the AISC Specification, select a WT beam with a 5-in. nominal depth for span and uniform dead and live loads as shown in Figure F.10. The toe of the stem of the WT is in tension. There is no deflection limit for this member. The beam is simply supported and continuously braced. The WT is ASTM A992 material.

Fig. F.10. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wu  1.2  0.08 kip/ft   1.6  0.24 kip/ft 

 0.320 kip/ft

 0.480 kip/ft From AISC Manual Table 3-23, Case 1: Mu  

wu L2 8

 0.480 kip/ft  6 ft 2

8  2.16 kip-ft

ASD wa  0.08 kip/ft  0.24 kip/ft

From AISC Manual Table 3-23, Case 1: Ma  

wa L2 8

 0.320 kip/ft  6 ft 2

8  1.44 kip-ft

Try a WT56. From AISC Manual Table 1-8, the geometric properties are as follows: WT56

d = 4.94 in. Ix = 4.35 in.4 Zx = 2.20 in.3 Sx = 1.22 in.3 bf = 3.96 in. tf = 0.210 in. y = 1.36 in.

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bf/2tf = 9.43 S xc  

Ix y 4.35 in.4 1.36 in.

 3.20 in.3

Nominal Flexural Strength Yielding From AISC Specification Section F9.1, for the limit state of yielding: Mn  M p

(Spec. Eq. F9-1)

M y  Fy S x

(Spec. Eq. F9-3)



  50 ksi  1.22 in.3



 61.0 kip-in. M p  Fy Z x  1.6 M y (for stems in tension)



  50 ksi  2.20 in.

3

(Spec. Eq. F9-2)

  1.6  61.0 kip-in.

 110 kip-in.  97.6 kip-in.  97.6 kip-in. or 8.13 kip-ft

Lateral-Torsional Buckling From AISC Specification Section F9.2, because the WT is continuously braced, the limit state of lateral-torsional buckling does not apply. Flange Local Buckling The limit state of flange local buckling is checked using AISC Specification Section F9.3. Flange Slenderness



bf 2t f

 9.43 From AISC Specification Table B4.1b, Case 10, the limiting width-to-thickness ratio for the flange is:  pf  0.38  0.38

E Fy 29,000 ksi 50 ksi

 9.15

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 rf  1.0  1.0

E Fy 29,000 ksi 50 ksi

 24.1

Because  pf     rf , the flange is noncompact and the limit state of flange local buckling will apply. From AISC Specification Section F9.3, the nominal flexural strength of a tee with a noncompact flange is:

     pf   M n   M p   M p  0.7 Fy S xc      1.6M y    rf   pf     9.43  9.15    110 kip-in.  110 kip-in.  0.7  50 ksi  3.20 in.3      97.6 kip-in.    24.1  9.15     110 kip-in.  97.6 kip-in.





 97.6 kip-in. Flexural yielding controls: M n  97.6 kip-in. or 8.13 kip-ft Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: LRFD

ASD

b  0.90

b  1.67

b M n  0.90  8.13 kip-ft 

M n 8.13 kip-ft  b 1.67  4.87 kip-ft  1.44 kip-ft o.k.

 7.32 kip-ft  2.16 kip-ft o.k.

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(Spec. Eq. F9-14)

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EXAMPLE F.11A

SINGLE-ANGLE FLEXURAL MEMBER WITH BRACING AT ENDS ONLY

Given:

Directly applying the requirements of the AISC Specification, select a single angle for span and uniform dead and live loads as shown in Figure F.11A. The vertical leg of the single angle is up and the toe is in compression. There are no horizontal loads. There is no deflection limit for this angle. The beam is simply supported and braced at the end points only. Assume bending about the geometric x-x axis and that there is no lateral-torsional restraint. The angle is ASTM A36 material.

Fig. F.11A. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wux  1.2  0.05 kip/ft   1.6  0.15 kip/ft 

 0.200 kip/ft

 0.300 kip/ft From AISC Manual Table 3-23, Case 1: M ux 



wux L2 8

 0.300 kip/ft  6 ft 2

8  1.35 kip-ft

ASD wax  0.05 kip/ft  0.15 kip/ft

From AISC Manual Table 3-23, Case 1: M ax 



wax L2 8

 0.200 kip/ft  6 ft 2

8  0.900 kip-ft

Try a L444. From AISC Manual Table 1-7, the geometric properties are as follows: L444

Sx = 1.03 in.3 Nominal Flexural Strength Yielding

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From AISC Specification Section F10.1, the nominal flexural strength due to the limit state of flexural yielding is: (Spec. Eq. F10-1)

M n  1.5 M y  1.5 Fy S x



 1.5  36 ksi  1.03 in.3



 55.6 kip-in.

Lateral-Torsional Buckling From AISC Specification Section F10.2, for single angles bending about a geometric axis with no lateral-torsional restraint, My is taken as 0.80 times the yield moment calculated using the geometric section modulus. M y  0.80 Fy S x



 0.80  36 ksi  1.03 in.3



 29.7 kip-in.

Determine Mcr. For bending moment about one of the geometric axes of an equal-leg angle with no axial compression, with no lateral-torsional restraint, and with maximum compression at the toe, use AISC Specification Equation F10-5a. Cb = 1.14 from AISC Manual Table 3-1

M cr 

2  0.58Eb4tCb   Lb t   1  0.88  1  2    Lb2 b   

(Spec. Eq. F10-5a)

2   4   6 ft 12 in./ft 4 in.  0.58  29,000 ksi  4.00 in. 4 in.1.14        1  1  0.88 2 2  4.00 in.      6 ft 12 in./ft     107 kip-in.

M y 29.7 kip-in.  ; M cr 107 kip-in.  0.278  1.0; therefore, AISC Specification Equation F10-2 is applicable

 My  M n  1.92  1.17  M y  1.5M y  M cr    29.7 kip-in.   1.92  1.17   29.7 kip-in.  1.5  29.7 kip-in. 107 kip-in.    38.7 kip-in.  44.6 kip-in.  38.7 kip-in. Leg Local Buckling AISC Specification Section F10.3 applies when the toe of the leg is in compression.

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Check slenderness of the leg in compression.

b t 4.00 in. = 4 in.  16.0

=

Determine the limiting compact slenderness ratios from AISC Specification Table B4.1b, Case 12. E Fy

 p = 0.54

29,000 ksi 36 ksi

= 0.54  15.3

Determine the limiting noncompact slenderness ratios from AISC Specification Table B4.1b, Case 12. E Fy

 r = 0.91 = 0.91

29,000 ksi 36 ksi

 25.8  p <  <  r , therefore, the leg is noncompact in flexure

Sc  0.80S x



 0.80 1.03in.3



 0.824 in.3   b  Fy  M n  Fy Sc  2.43  1.72     t  E  

(Spec. Eq. F10-6)

 36 ksi    36 ksi  0.824 in.3  2.43  1.72 16.0   29, 000 ksi    43.3 kip-in.





The lateral-torsional buckling limit state controls. Mn = 38.7 kip-in. or 3.23 kip-ft Available Flexural Strength From AISC Specification Section F1, the available flexural strength is:

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ASD

LRFD b  0.90

b  1.67

b M n  0.90  3.23 kip-ft 

M n 3.23kip-ft  b 1.67  1.93 kip-ft  0.900 kip-ft o.k.

 2.91 kip-ft  1.35 kip-ft o.k.

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EXAMPLE F.11B SINGLE-ANGLE FLEXURAL MEMBER WITH BRACING AT ENDS AND MIDSPAN Given:

Directly applying the requirements of the AISC Specification, select a single angle for span and uniform dead and live loads as shown in Figure F.11B. The vertical leg of the single angle is up and the toe is in compression. There are no horizontal loads. There is no deflection limit for this angle. The beam is simply supported and braced at the end points and midspan. Assume bending about the geometric x-x axis and that there is lateral-torsional restraint at the midspan and ends only. The angle is ASTM A36 material.

Fig. F.11B. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wux  1.2  0.05 kip/ft   1.6  0.15 kip/ft 

 0.200 kip/ft

 0.300 kip/ft From AISC Manual Table 3-23, Case 1: M ux 



wux L2 8

 0.300 kip/ft  6 ft 2

8  1.35 kip-ft

ASD wax  0.05 kip/ft  0.15 kip/ft

From AISC Manual Table 3-23, Case 1: M ax 



wax L2 8

 0.200 kip/ft  6 ft 2

8  0.900 kip-ft

Try a L444. From AISC Manual Table 1-7, the geometric properties are as follows: L444

Sx = 1.03 in.3

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Nominal Flexural Strength Flexural Yielding From AISC Specification Section F10.1, the nominal flexural strength due to the limit state of flexural yielding is: (Spec. Eq. F10-1)

M n  1.5 M y  1.5 Fy S x



 1.5  36 ksi  1.03 in.3



 55.6 kip-in.

Lateral-Torsional Buckling From AISC Specification Section F10.2(b)(2)(ii), for single angles with lateral-torsional restraint at the point of maximum moment, My is taken as the yield moment calculated using the geometric section modulus. M y  Fy S x



  36 ksi  1.03 in.3



 37.1 kip-in.

Determine Mcr. For bending moment about one of the geometric axes of an equal-leg angle with no axial compression, with lateraltorsional restraint at the point of maximum moment only (at midspan in this case), and with maximum compression at the toe, Mcr shall be taken as 1.25 times Mcr computed using AISC Specification Equation F10-5a. Cb = 1.30 from AISC Manual Table 3-1

 0.58Eb4tCb M cr  1.25  Lb 2 

2    Lb t    1  0.88  1   2   b    

(from Spec. Eq. F10-5a)

2   0.58  29, 000 ksi  4.00 in.4 4 in.1.30      3 ft 12 in./ft 4 in.        1  1.25  1  0.88  2 2    4.00 in.     3 ft 12 in./ft       176 kip-in.

My M cr

37.1 kip-in. 176 kip-in.  0.211  1.0; therefore, AISC Specification Equation F10-2 is applicable 

 My  M n  1.92  1.17  M y  1.5M y  M cr    37.1 kip-in.   1.92  1.17   37.1 kip-in.  1.5  37.1kip-in. 176 kip-in.    51.3 kip-in.  55.7 kip-in.  51.3 kip-in.

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(Spec. Eq. F10-2)

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Leg Local Buckling Mn = 43.3 kip-in. from Example F.11A. The leg local buckling limit state controls. Mn = 43.3 kip-in. or 3.61 kip-ft Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: ASD

LRFD b  0.90

b  1.67

b M n  0.90  3.61 kip-ft 

M n 3.61 kip-ft  b 1.67  2.16 kip-ft  0.900 kip-ft o.k.

 3.25 kip-ft  1.35 kip-ft o.k.

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EXAMPLE F.11C

SINGLE-ANGLE FLEXURAL MEMBER WITH VERTICAL AND HORIZONTAL LOADING

Given: Directly applying the requirements of the AISC Specification, select a single angle for span and uniform vertical dead and live loads as shown in Figure F.11C-1. The horizontal load is a uniform wind load. There is no deflection limit for this angle. The angle is simply supported and braced at the end points only and there is no lateral-torsional restraint. Use load combination 4 from Section 2.3.1 of ASCE/SEI 7 for LRFD and load combination 6 from Section 2.4.1 of ASCE/SEI 7 for ASD. The angle is ASTM A36 material.

(a) Beam bracing diagram

(b) Beam loading

Fig. F.11C-1. Beam loading and bracing diagram.

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wux  1.2  0.05 kip/ft   0.15 kip/ft

 0.210 kip/ft wuy  1.0  0.12 kip/ft   0.120 kip/ft M ux 



wux L2 8

 0.210 kip/ft  6 ft 2

8  0.945 kip-ft

ASD wax  0.05 kip/ft  0.75  0.15 kip/ft 

 0.163 kip/ft way  0.75  0.6  0.12 kip/ft    0.0540 kip/ft

M ax 



wax L2 8

 0.163 kip/ft  6 ft 2

8  0.734 kip-ft

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LRFD

ASD

2

M uy 



2

wuy L 8

M ay 

 0.120 kip/ft  6 ft 2



8  0.540 kip-ft

way L 8

 0.0540 kip/ft  6 ft 2

8  0.243 kip-ft

Try a L444. Sign convention for geometric axes moments are: ASD

LRFD Mux = 0.945 kip-ft

Max = 0.734 kip-ft

Muy = 0.540 kip-ft

May = 0.243 kip-ft

As shown in Figure F.11C-2, the principal axes moments are: ASD M aw  M ax cos   M ay sin 

LRFD M uw  M ux cos   M uy sin 

  0.945 kip-ft  cos 45 

  0.734 kip-ft  cos 45 

  0.540 kip-ft  sin 45 

  0.243 kip-ft  sin 45 

 0.286 kip-ft

 0.347 kip-ft

M uz   M ux sin   M uy cos 

M az   M ax sin   M ay cos 

   0.945 kip-ft  sin 45 

   0.734 kip-ft  sin 45 

  0.540 kip-ft  cos 45 

  0.243 kip-ft  cos 45 

 1.05 kip-ft

 0.691 kip-ft

(a) Positive geometric and principal axes

(b) Principal axis moments

Fig. F.11C-2. Example F.11C single angle geometric and principal axes moments.

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From AISC Manual Table 1-7, the geometric properties are as follows: L444

A = 1.93 in.2 Sx= Sy = 1.03 in.3 Ix = Iy = 3.00 in.4 Iz = 1.19 in.4 rz = 0.783 in. Additional principal axes properties from the AISC Shapes Database are as follows: wB wC zC Iw SzB SzC SwC

= 1.53 in. = 1.39 in. = 2.74 in. = 4.82 in.4 = 0.778 in.3 = 0.856 in.3 = 1.76 in.3

Z-Axis Nominal Flexural Strength Note that Muz and Maz are positive; therefore, the toes of the angle are in compression. Flexural Yielding From AISC Specification Section F10.1, the nominal flexural strength due to the limit state of flexural yielding is: (from Spec. Eq. F10-1)

M nz  1.5 M y  1.5 Fy S zB



 1.5  36 ksi  0.778 in.3



 42.0 kip-in.

Lateral-Torsional Buckling From the User Note in AISC Specification Section F10, the limit state of lateral-torsional buckling does not apply for bending about the minor axis. Leg Local Buckling Check slenderness of outstanding leg in compression.

b t 4.00 in. = 4 in.  16.0

=

From AISC Specification Table B4.1b, Case 12, the limiting width-to-thickness ratios are:

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 p = 0.54 = 0.54

E Fy 29,000 ksi 36 ksi

 15.3

 r = 0.91 = 0.91

E Fy

29,000 ksi 36 ksi

 25.8

Because  p <  <  r , the leg is noncompact in flexure.

Sc  S zC (to toe in compression)  0.856 in.3   b  Fy  M nz = Fy Sc  2.43  1.72     t  E  

(Spec. Eq. F10-6)

 36 ksi  =  36 ksi  0.856 in.3  2.43  1.72 16.0   29, 000 ksi    45.0 kip-in.





The flexural yielding limit state controls. Mnz = 42.0 kip-in. or 3.50 kip-ft Z-Axis Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: ASD

LRFD

b  0.90

b  1.67

b M nz  0.90  3.50 kip-ft 

M nz 3.50 kip-ft  b 1.67  2.10 kip-ft

 3.15 kip-ft W-Axis Nominal Flexural Strength Flexural Yielding

(from Spec. Eq. F10-1)

M nw  1.5 M y  1.5 Fy S wC



 1.5  36 ksi  1.76 in.3



 95.0 kip-in.

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Lateral-Torsional Buckling Determine Mcr. For bending about the major principal axis of an equal-leg angle without continuous lateral-torsional restraint, use AISC Specification Equation F10-4. Cb = 1.14 from Manual Table 3-1 From AISC Specification Section F10.2(b)(1), w  0 for equal leg angles.

M cr 



2   9EArz tCb   r   r  1   4.4 w z   4.4 w z  8Lb  Lb t  Lb t    



(Spec. Eq. F10-4)



9  29, 000 ksi  1.93 in.2  0.783 in.4 in.1.14  8  6 ft 12 in./ft 

2       0  0.783 in. 0  0.783 in.    1   4.4  4.4      6 ft 12 in./ft 4 in.    6 ft 12 in./ft 4 in.       195 kip-in.

M y  Fy S wC



  36 ksi  1.76 in.3



 63.4 kip-in.

M y 63.4 kip-in.  M cr 195 kip-in.  0.325  1.0, therefore, AISC Specification Equation F10-2 is applicable

 My  M nw  1.92  1.17  M y  1.5M y  M cr    63.4 kip-in.   1.92  1.17   63.4 kip-in.  1.5  63.4 kip-in. 195 kip-in.    79.4 kip-in.  95.1 kip-in.  79.4 kip-in. Leg Local Buckling From the preceding calculations, the leg is noncompact in flexure.

Sc  S wC (to toe in compression)  1.76 in.3

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  b  Fy  M nw  Fy Sc  2.43  1.72     t  E  

(Spec. Eq. F10-6)

 36 ksi  =  36 ksi  1.76 in.3  2.43  1.72 16.0   29, 000 ksi    92.5 kip-in.





The lateral-torsional buckling limit state controls. Mnw = 79.4 kip-in. or 6.62 kip-ft W-Axis Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: ASD

LRFD b  0.90

b  1.67

b M nw  0.90  6.62 kip-ft 

M nw 6.62 kip-ft  b 1.67  3.96 kip-ft

 5.96 kip-ft Combined Loading

The moment resultant has components about both principal axes; therefore, the combined stress ratio must be checked using the provisions of AISC Specification Section H2. f ra f f  rbw  rbz  1.0 Fca Fcbw Fcbz

(Spec. Eq. H2-1)

Note: Rather than convert moments into stresses, it is acceptable to simply use the moments in the interaction equation because the section properties that would be used to convert the moments to stresses are the same in the numerator and denominator of each term. It is also important for the designer to keep track of the signs of the stresses at each point so that the proper sign is applied when the terms are combined. The sign of the moments used to convert geometric axis moments to principal axis moments will indicate which points are in tension and which are in compression but those signs will not be used in the interaction equations directly. Based on Figure F.11C-2, the required flexural strength and available flexural strength for this beam can be summarized as: ASD

LRFD

M uw  0.286 kip-ft

M aw  0.347 kip-ft

b M nw  5.96 kip-ft

M nw  3.96 kip-ft b

M uz  1.05 kip-ft

M az  0.691 kip-ft

b M nz  3.15 kip-ft

M nz  2.10 kip-ft b

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At point B: Mw causes no stress at point B; therefore, the stress ratio is set to zero. Mz causes tension at point B; therefore it will be taken as negative. ASD

LRFD 0

1.05 kip-ft  0.333  1.0 3.15 kip-ft

0

o.k.

0.691 kip-ft  0.329  1.0 2.10 kip-ft

o.k.

At point C: Mw causes tension at point C; therefore, it will be taken as negative. Mz causes compression at point C; therefore, it will be taken as positive. LRFD 0.286 kip-ft 1.05 kip-ft    0.285  1.0 o.k. 5.96 kip-ft 3.15 kip-ft

ASD 0.347 kip-ft 0.691 kip-ft    0.241  1.0 o.k. 3.96 kip-ft 2.10 kip-ft

At point A: Mw and Mz cause compression at point A; therefore, both will be taken as positive. LRFD 0.286 kip-ft 1.05 kip-ft   0.381  1.0 o.k. 5.96 kip-ft 3.15 kip-ft

ASD 0.347 kip-ft 0.691 kip-ft   0.417  1.0 o.k. 3.96 kip-ft 2.10 kip-ft

Thus, the interaction of stresses at each point is seen to be less than 1.0 and this member is adequate to carry the required load. Although all three points were checked, it was expected that point A would be the controlling point because compressive stresses add at this point.

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EXAMPLE F.12 RECTANGULAR BAR IN MAJOR AXIS BENDING Given:

Directly applying the requirements of the AISC Specification, select a rectangular bar for span and uniform vertical dead and live loads as shown in Figure F.12. The beam is simply supported and braced at the end points and midspan. Conservatively use Cb = 1.0. Limit the depth of the member to 5 in. The bar is ASTM A36 material.

Fig. F.12. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-5, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wu  1.2  0.44 kip/ft   1.6 1.32 kip/ft 

 1.76 kip/ft

 2.64 kip/ft From AISC Manual Table 3-23, Case 1: Mu 



ASD wa  0.44 kip/ft  1.32 kip/ft

wu L2 8

From AISC Manual Table 3-23, Case 1: Ma 

 2.64 kip/ft 12 ft 2

8  47.5 kip-ft



wa L2 8

1.76 kip/ft 12 ft 2

8  31.7 kip-ft

Try a BAR 5 in. 3 in. From AISC Manual Table 17-27, the geometric properties are as follows: Sx  

bd 2 6

 3.00 in. 5.00 in.2 6

 12.5 in.3

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Zx  

bd 2 4

 3.00 in. 5.00 in.2 4 3

 18.8 in.

Nominal Flexural Strength Flexural Yielding Check limit from AISC Specification Section F11.1. Lb d t

2



 6 ft 12 in./ft  5.00 in.  3.00 in.2

 40.0 0.08 E 0.08  29, 000 ksi   36 ksi Fy  64.4  40.0; therefore, the yielding limit state applies M n  M p  Fy Z  1.6 Fy S

1.6 Fy S  1.6 Fy S x



 1.6  36 ksi  12.5 in.3

(Spec. Eq. F11-1)



 720 kip-in. Fy Z  Fy Z x



  36 ksi  18.8 in.3



 677 kip-in.  720 kip-in.

Use Mn = 677 kip-in. or 56.4 kip-ft. Lateral-Torsional Buckling From AISC Specification Section F11.2(a), because

Lb d t

2



0.08E , the lateral-torsional buckling limit state does not Fy

apply. Available Flexural Strength From AISC Specification Section F1, the available flexural strength is:

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ASD

LRFD b = 0.90

b = 1.67

b M n  0.90  56.4 kip-ft 

M n 56.4 kip-ft  b 1.67  33.8 kip-ft  31.7 kip-ft o.k.

 50.8 kip-ft  47.5 kip-ft o.k.

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EXAMPLE F.13 ROUND BAR IN BENDING Given:

Select a round bar for span and concentrated dead and live loads, at midspan, as shown in Figure F.13. The beam is simply supported and braced at the end points only. Conservatively use Cb = 1.0. Limit the diameter of the member to 2 in. The weight of the bar is negligible. The bar is ASTM A36 material.

Fig. F.13. Beam loading and bracing diagram.

Solution:

From AISC Manual Table 2-5, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From Chapter 2 of ASCE/SEI 7 the required flexural strength is: ASD

LRFD Pu  1.2  0.10 kip   1.6  0.25 kip 

Pa  0.10 kip  0.25 kip  0.350 kip

 0.520 kip From AISC Manual Table 3-23, Case 7: Mu  

Pu L 4 0.520 kip  2.5 ft  

4  0.325 kip-ft

From AISC Manual Table 3-23, Case 7: Ma  

Pa L 4  0.350 kip  2.5 ft 

4  0.219 kip-ft

Try a BAR 1-in.-diameter. From AISC Manual Table 17-27, the geometric properties are as follows: S

d 3 32  1.00 in.

3



32  0.0982 in.3

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Z 

d3 6

1.00 in.3

6  0.167 in.3 Nominal Flexural Strength Flexural Yielding From AISC Specification Section F11.1, the nominal flexural strength based on the limit state of flexural yielding is: M n  M p  Fy Z  1.6 Fy S x



1.6 Fy S  1.6  36 ksi  0.0982 in.3

(Spec. Eq. F11-1)



 5.66 kip-in.



Fy Z   36 ksi  0.167 in.3



 6.01 kip-in.  5.66 kip-in, therefore, M n  5.66 kip-in.

From AISC Specification Section F11.2, the limit state lateral-torsional buckling need not be considered for rounds. The flexural yielding limit state controls. Mn = 5.66 kip-in. or 0.472 kip-ft Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: ASD

LRFD b = 1.67

b = 0.90

b M n  0.90  0.472 kip-ft   0.425 kip-ft  0.325 kip-ft o.k.



M n 0.472 kip-ft  b 1.67   0.283 kip-ft  0.219 kip-ft o.k.

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EXAMPLE F.14 POINT-SYMMETRICAL Z-SHAPE IN MAJOR AXIS BENDING Given:

Directly applying the requirements of the AISC Specification, determine the available flexural strength of a Zshaped flexural member for the span and loading shown in Figure F.14-1. The beam is simply supported and braced at the third and end points. Assume Cb = 1.0. Assume the beam is loaded through the shear center. The geometry for the member is shown in Figure F.14-2. The member is ASTM A36 material.

Fig. F.14-1. Beam loading and bracing diagram.

Fig. F.14-2. Beam geometry for Example F.14.

Solution:

From AISC Manual Table 2-5, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi

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The geometric properties are as follows:

tw  t f  4 in.

A  2  2.50 in.4 in.  2 4 in.4 in.  11.5 in.4 in.  4.25 in.2  4 in.4 in.3    2.50 in.4 in.3  2 2 2  4 in.  5.63 in.   2    2.50 in.4 in. 5.88 in.  Ix  2  12 12     +

4 in.11.5 in.3 12

 78.9 in.4

y  6.00 in. Sx 

Ix y

78.9 in.4 6.00 in.  13.2 in.3 

 4 in.4 in.3   4 in. 2.50 in.3  2 2 2  4 in.  2.25 in.   2    2.50 in.4 in.1.13 in.  Iy  2 12 12     +

11.5 in.4 in.3 12

 2.90 in.4 ry  

Iy A 2.90 in.4

4.25 in.2  0.826 in.

The effective radius of gyration, rts, may be conservatively approximated from the User Note in AISC Specification Section F2.2. A more exact method may be derived as discussed in AISC Design Guide 9, Torsional Analysis of Structural Steel Members (Seaburg and Carter, 1997), for a Z-shape that excludes lips. From AISC Specification Section F2.2 User Note:

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bf

rts 

 1 htw  12 1    6 bf t f  2.50 in.



  1   11.5 in.4 in.   12 1        6    2.50 in.4 in.  

 0.543 in. From Chapter 2 of ASCE/SEI 7, the required flexural strength is: LRFD wu  1.2  0.025 kip/ft   1.6  0.10 kip/ft 

 0.125 kip/ft

 0.190 kip/ft From AISC Manual Table 3-23, Case 1: Mu  

ASD wa  0.025 kip/ft  0.10 kip/ft

wu L2 8

From AISC Manual Table 3-23, Case 1: Ma 

 0.190 kip/ft 18 ft 2



wa L2 8

 0.125 kip/ft 18 ft 2

8  5.06 kip-ft

8  7.70 kip-ft

Nominal Flexural Strength Flexural Yielding From AISC Specification Section F12.1, the nominal flexural strength based on the limit state of flexural yielding is, Fn  Fy

(Spec. Eq. F12-2)

 36 ksi

M n  Fn Smin



  36 ksi  13.2 in.

3

(Spec. Eq. F12-1)



 475 kip-in.

Local Buckling There are no specific local buckling provisions for Z-shapes in the AISC Specification. Use provisions for rolled channels from AISC Specification Table B4.1b, Cases 10 and 15. Flange Slenderness Conservatively neglecting the end return,

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

b tf

2.50 in. 4 in.  10.0 

E Fy

 p  0.38  0.38

(Spec. Table B4.1b, Case 10)

29, 000 ksi 36 ksi

 10.8    p ; therefore, the flange is compact

Web Slenderness

h tw 11.5 in.  4 in.  46.0



 p  3.76  3.76

E Fy

(Spec. Table B4.1b, Case 15)

29, 000 ksi 36 ksi

 107    p ; therefore, the web is compact

Therefore, the local buckling limit state does not apply. Lateral-Torsional Buckling Per the User Note in AISC Specification Section F12, take the critical lateral-torsional buckling stress as half that of the equivalent channel. This is a conservative approximation of the lateral-torsional buckling strength which accounts for the rotation between the geometric and principal axes of a Z-shaped cross section, and is adopted from the North American Specification for the Design of Cold-Formed Steel Structural Members (AISI, 2016). Calculate limiting unbraced lengths. For bracing at 6 ft on center,

Lb   6 ft 12 in./ft   72.0 in.

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E Fy

L p  1.76ry

(Spec. Eq. F2-5)

29, 000 ksi 36 ksi  41.3 in.  72.0 in.  1.76  0.826 in.

Per the User Note in AISC Specification Section F2, the square root term in AISC Specification Equation F2-4 can conservatively be taken equal to one. Therefore, Equation F2-6 can also be simplified. Substituting 0.7Fy for Fcr (where Fcr is half of the critical lateral-torsional buckling stress of the equivalent channel) in Equation F2-4 and solving for Lb = Lr, AISC Specification Equation F2-6 becomes: Lr  rts

0.5 E 0.7 Fy

   0.543 in.

0.5  29, 000 ksi  0.7  36 ksi 

 40.9 in.  72.0 in.

Calculate one half of the critical lateral-torsional buckling stress of the equivalent channel. Lb > Lr, therefore, Fcr   0.5 

Cb 2 E  Lb  r   ts 

2

 Jc   Lb  1  0.078     S x ho   rts 

2

(from Spec. Eq. F2-4)

Conservatively taking the square root term as 1.0,  C 2 E  Fcr   0.5   b 2  1.0    Lb    r     ts   1.0 2  29, 000 ksi     0.5    1.0  2   72.0 in.     0.543 in.        8.14 ksi Fn  Fcr  Fy

(Spec. Eq. F12-3)

 8.14 ksi  36 ksi

M n  Fn Smin



o.k.

  8.14 ksi  13.2 in.3

(Spec. Eq. F12-1)



 107 kip-in.

The lateral-torsional buckling limit state controls. Mn = 107 kip-in. or 8.92 kip-ft

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Available Flexural Strength From AISC Specification Section F1, the available flexural strength is: LRFD

ASD b = 1.67

b = 0.90

b M n  0.90  8.92 kip-ft   8.03 kip-ft  7.70 kip-ft o.k.



M n 8.92 kip-ft  b 1.67  5.34 kip-ft  5.06 kip-ft o.k.

Because the beam is loaded through the shear center, consideration of a torsional moment is unnecessary. If the loading produced torsion, the torsional effects should be evaluated using AISC Design Guide 9, Torsional Analysis of Structural Steel Members (Seaburg and Carter, 1997).

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EXAMPLE F.15 PLATE GIRDER FLEXURAL MEMBER Given:

Verify the built-up plate girder for the span and loads as shown in Figure F.15-1 with a cross section as shown in Figure F.15-2. The beam has a concentrated dead and live load at midspan and a uniformly distributed self weight. The plate girder is simply supported and is laterally braced at quarter and end points. The deflection of the girder is limited to 1 in. The plate girder is ASTM A572 Grade 50 material. The flange-to-web welds will be designed for both continuous and intermittent fillet welds using 70-ksi electrodes.

Fig. F.15-1. Beam loading and bracing diagram.

Fig. F.15-2. Plate girder geometry. Solution:

From AISC Manual Table 2-5, the material properties are as follows: ASTM A572 Grade 50 Fy = 50 ksi Fu = 65 ksi

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From ASCE/SEI 7, Chapter 2, the required shear and flexural strengths are: LRFD Pu  1.2  240 kips   1.6 160 kips 

ASD

Pa  240 kips  160 kips  400 kips

 544 kips wu  1.2  0.296 kip/ft 

wa  0.296 kip/ft

 0.355 kip/ft

Pa wa L  2 2 400 kips  0.296 kip/ft  50 ft    2 2  207 kips

Pu wu L  2 2 544 kips  0.355 kip/ft  50 ft    2 2  281 kips

Va 

Vu 

Mu  

Pu L wu L2  4 8

Ma 

 544 kips  50 ft   0.355 kip/ft  50 ft 2 

4  6,910 kip-ft

8



Pa L wa L2  4 8

 400 kips  50 ft   0.296 kip/ft  50 ft 2

4  5, 090 kip-ft



8

Proportioning Limits The proportioning limits from AISC Specification Section F13.2 are evaluated as follows, where a is the clear distance between transverse stiffeners. a  25 ft 12 in./ft   62 in. h  4.84

Because a h  1.5, use AISC Specification Equation F13-4. 0.40 E h  t  Fy  w  max 

(Spec. Eq. F13-3)

0.40  29, 000 ksi  50 ksi

 232 h 62 in.  tw 2 in.  124  232 o.k.

From AISC Specification Section F13.2, the following limit applies to all built-up I-shaped members: hc tw  62 in.2 in.   10 bf t f 14 in. 2 in.  1.11  10

o.k.

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Section Properties

bh3   Ad 2 12

Ix   

2 in. 62 in.3 12

 14 in. 2 in.3  2   2  2 in.14 in. 32.0 in.   2   12  

 67,300 in.4 S xt  S xc  

Ix

 d 2 67,300 in.4  66 in. 2 

 2, 040 in.3 Z x   Ay

  2 2 in. 31.0 in. 31.0 in. 2    2  2 in.14 in. 32.0 in.  2, 270 in.3

J 

bt 3 3

 14 in. 2 in.3   62 in.2 in.3   2 3 3    77.3 in.4

ho  h  t f  62 in.  2 in.  64.0 in. Deflection The maximum deflection is:  

 PD  PL  L3 48 EI



5wD L4 384 EI

 240 kips  160 kips  50 ft 3 12 in./ft 3 5  0.296 kip/ft  50 ft 4 12 in./ft 3  48  29, 000 ksi   67,300 in.4  384  29, 000 ksi   67,300 in.4 

 0.944in.  1.00in. o.k. Web Slenderness

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h tw 62 in.  2 in.  124



The limiting width-to-thickness ratios for the web are:  pw  3.76  3.76

E from AISC Specification Table B4.1b, Case 15 Fy 29, 000 ksi 50 ksi

 90.6  rw  5.70  5.70

E from AISC Specification Table B4.1b, Case 15 Fy 29, 000 ksi 50 ksi

 137  pw     rw , therefore the web is noncompact and AISC Specification Section F4 applies.

Flange Slenderness   

b t bf 2t f 14 in. 2  2 in.

 3.50

 pf  0.38

E from AISC Specification Table B4.1b, Case 11 Fy

29, 000 ksi 50 ksi  9.15  , therefore the flanges are compact  0.38

Nominal Flexural Strength Compression Flange Yielding The web plastification factor is determined using AISC Specification Section F4.2(c)(6).

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I yc  

t f bf 3 12

 2 in.14 in.3

12  457 in.4  t f b f 3  htw3 I y  2   12  12     2 in.14 in.3   62 in.2 in.3   2 12 12    915 in.4

I yc 457 in.4  I y 915 in.4  0.499 Because Iyc/Iy > 0.23, AISC Specification Section F4.2(c)(6)(i) applies. M p  Fy Z x  1.6 Fy S x









  50 ksi  2, 270 in.3 1 ft/12 in.  1.6  50 ksi  2, 040 in.3 1 ft/12 in.  9, 460 kip-ft  13, 600 kip-ft  9, 460 kip-ft M yc  Fy S xc

(Spec. Eq. F4-4)

  50 ksi  2, 040 kip-in.1 ft/12 in.  8,500 kip-ft

hc  h  62 in. hc tw 62 in.  2 in.  124   pw  90.6; therefore use AISC Specification Equation F4-9b



R pc 



     pw  M p Mp  Mp    1    M yc  M yc    rw   pw  M yc 9, 460 kip-ft  9, 460 kip-ft   124  90.6  9, 460 kip-ft   1   8,500 kip-ft  8,500 kip-ft   137  90.6  8,500 kip-ft

 1.03  1.11  1.03

The nominal flexural strength is calculated as:

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M n  R pc M yc

(Spec. Eq. F4-1)

 1.03 8,500 kip-ft   8, 760 kip-ft

From AISC Specification Section F4.1, the available flexural strength is: LRFD

ASD

b  0.90

b  1.67

b M n  0.90  8, 760 kip-ft 

M n 8, 760 kip-ft  b 1.67  5, 250 kip-ft  5, 090 kip-ft o.k.

 7,880 kip-ft  6,910 kip-ft o.k.

Lateral-Torsional Buckling The middle unbraced lengths control by inspection. For bracing at quarter points,

Lb  12.5 ft 12 in./ft   150 in. aw  

hc tw b fc t fc

(Spec. Eq. F4-12)

 62 in.2 in. 14 in. 2 in.

 1.11

rt 



b fc

(Spec. Eq. F4-11)

 1  12  1  aw   6  14.0 in.

  1.11   12 1      6   3.71 in.

From AISC Specification Equation F4-7: L p  1.1rt

E Fy

(Spec. Eq. F4-7)

29, 000 ksi 50 ksi  98.3  150 in.; therefore, lateral-torsional buckling applies  1.1 3.71 in.

From AISC Specification Section F4.2(c)(3):

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S xt 2, 040 in.3  S xc 2, 040 in.3  1.00  0.7; therefore, AISC Specification Equation F4-6a applies

FL  0.7 Fy

(Spec. Eq. F4-6a)

 0.7  50 ksi   35.0 ksi From AISC Specification Equation F4-8:

Lr  1.95rt

E FL

2

J  J   FL      6.76  E  S xc ho S h    xc o 

 29, 000 ksi   1.95  3.71 in.    35.0 ksi 



2

(Spec. Eq. F4-8) 2

2   77.3 in.4  35.0 ksi     6.76    3  29, 000 ksi  2, 040 in.3  64.0 in.  2, 040 in.  64.0 in. 

77.3 in.4







 369 in.

L p  Lb  Lr ; therefore, use AISC Specification Equation F4-2

The lateral-torsional buckling modification factor is determined by solving for the moment in the beam using statics. Note: The following solution uses LRFD load combinations. Using ASD load combinations will give approximately the same solution for Cb. M max  6,910 kip-ft M A  4, 350 kip-ft M B  5, 210 kip-ft M C  6, 060 kip-ft

Cb  

12.5M max 2.5M max  3M A  4 M B  3M C

(Spec. Eq. F1-1)

12.5  6,910 kip-ft 

2.5  6,910 kip-ft   3  4,350 kip-ft   4  5, 210 kip-ft   3  6, 060 kip-ft 

 1.25

The nominal flexural strength is calculated as:   Lb  L p M n  Cb  R pc M yc   R pc M yc  FL S xc    Lr  L p 

    R pc M yc  

(Spec. Eq. F4-2)

  150 in.  98.3 in.    1.25 8,760 kip-ft  8,760 kip-ft   35.0 ksi  2, 040 in.3 1 ft/12 in.      8,760 kip-ft    369 in.  98.3 in.     10,300 kip-ft  8,760 kip-ft





 8,760 kip-ft

From AISC Specification Section F4.2, the available flexural strength is:

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LRFD

ASD

b  0.90

b  1.67

b M n  0.90  8, 760 kip-ft 

M n 8, 760 kip-ft  b 1.67  5, 250 kip-ft  5, 090 kip-ft o.k.

 7,880 kip-ft  6,910 kip-ft o.k.

Compression Flange Local Buckling From AISC Specification Section F4.3(a), this limit state does not apply because the flanges are compact. Tension Flange Yielding From AISC Specification Section F4.4(a), because S xt  S xc , this limit state does not apply. Nominal Shear Strength Determine the nominal shear strength without tension field action, using AISC Specification Section G2.1. For builtup I-shaped members, determine Cv1 and kv from AISC Specification Section G2.1(b). a  25.0 ft 12 in./ft   2 in.  62 in. h  4.83  3.0

From AISC Specification Section G2.1(b)(2): kv = 5.34 1.10

5.34  29, 000 ksi  kv E  1.10 Fy 50 ksi  61.2  h tw  124; therefore, AISC Specification Equation G2-4 applies

Cv1 

1.10 kv E Fy

(Spec. Eq. G2-4)

h tw

61.2 124  0.494 

The nominal shear strength is calculated as follows: Vn  0.6 Fy AwCv1

(Spec. Eq. G2-1)

 0.6  50 ksi  66 in.2 in. 0.494   489 kips

From AISC Specification Section G.1, the available shear strength is:

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LRFD

ASD

v  0.90

v  1.67

vVn  0.90  489 kips 

Vn 489 kips  v 1.67  293 kips  207 kips

 440 kips  281 kips o.k.

o.k.

Flange-to-Web Fillet Weld—Continuous Weld Calculate the required shear flow using VQ/Ix because the stress distribution is linearly elastic away from midspan. Q  Ay

 h tf   bf t f    2 2   62 in. 2 in.   14 in. 2 in.    2   2  896 in.3

LRFD

ASD

VQ Ru  u Ix 

VQ Ra  a Ix

 281 kips   896 in.3 



67,300 in.4  3.74 kip/in.

 207 kips   896 in.3 

67,300 in.4  2.76 kip/in.

From AISC Specification Table J2.4, the minimum fillet weld size that can be used on the 2-in.-thick web is:

wmin  x in. From AISC Manual Part 8, the required fillet weld size is: LRFD Dreq

Ru  1.392  2 sides  

ASD (from Manual Eq. 8-2a)

3.74 kip/in. 1.392  2 sides 

 1.34 sixteenths  3 sixteenths

Use w  x in.

Dreq

Ra  0.928  2 sides  

(from Manual Eq. 8-2b)

2.76 kip/in. 0.928  2 sides 

 1.49 sixteenths  3 sixteenths

Use w  x in.

From AISC Specification Equation J2-2, the available shear rupture strength of the web in kip/in. is:

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LRFD

  0.75

  2.00

ASD

Rn FnBM ABM    0.60 Fu tw   0.60  65 ksi 2 in.  2.00  9.75 kip/in.  2.76 kip/in. o.k.

Rn  FnBM ABM  0.60 Fu t w  0.75  0.60  65 ksi 2 in.  14.6 kip/in.  3.74 kip/in. o.k.

Flange-to-Web Fillet Weld—Intermittent Weld The two sided intermittent weld is designed using the minimum fillet weld size determined previously, wmin  x in., and spaced at 12 in. center-to-center. LRFD Ru  Rn

ASD (from Manual Eq. 8-2a)

 lreq   1.392 D  2 sides     s 

Solving for lreq, lreq  

Ru s 1.392 D  2 sides 

 3.74 kip-in.12 in. 1.392  3 sixteenth  2 sides 

 5.37 in. Use l = 6 in. at 12 in. o.c.

R Ru  n 

(from Manual Eq. 8-2b)

 lreq   0.928 D  2 sides     s 

Solving for lreq, lreq  

Ru s 0.928D  2 sides 

 2.76 kip-in.12 in. 0.928  3 sixteenth  2 sides 

 5.95 in. Use l = 6 in. at 12 in. o.c.

The limitations for a intermittent fillet weld are checked using AISC Specification Section J2.2b(e): l  4D 6 in.  4  x in. 6 in.  0.75 in. o.k. l  12 in. 6 in.  12 in. o.k.

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CHAPTER F DESIGN EXAMPLE REFERENCES AISI (2016), North American Specification for the Design of Cold-Formed Steel Structural Members, ANSI/AISI Standard S100, American Iron and Steel Institute, Washington D.C. Seaburg, P.A. and Carter, C.J. (1997), Torsional Analysis of Structural Steel Members, Design Guide 9, AISC, Chicago, IL.

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Chapter G Design of Members for Shear INTRODUCTION This Specification chapter addresses webs of singly or doubly symmetric members subject to shear in the plane of the web, single angles and HSS subject to shear, and shear in the weak direction of singly or doubly symmetric shapes. G1. GENERAL PROVISIONS The design shear strength, vVn, and the allowable shear strength, Vn /v, are determined as follows: Vn = nominal shear strength based on shear yielding or shear buckling v = 0.90 (LRFD)  v = 1.67 (ASD) Exception: For all current ASTM A6, W, S and HP shapes except W44230, W40149, W36135, W33118,

W3090, W2455, W1626 and W1214 for Fy = 50 ksi:

v = 1.00 (LRFD)  v = 1.50 (ASD) Strong axis shear values are tabulated for W-shapes in AISC Manual Tables 3-2, 3-6 and 6-2, for S-shapes in AISC Manual Table 3-7, for C-shapes in AISC Manual Table 3-8, and for MC-shapes in AISC Manual Table 3-9. Strong axis shear values are tabulated for rectangular HSS, round HSS and pipe in Part IV. Weak axis shear values for Wshapes, S-shapes, C-shapes and MC-shapes, and shear values for angles, rectangular HSS and box members are not tabulated. G2. I-SHAPED MEMBERS AND CHANNELS This section includes provisions for shear strength of webs without the use of tension field action and for interior web panels considering tension field action. Provisions for the design of transverse stiffeners are also included in Section G2. As indicated in the User Note of this section, virtually all W, S and HP shapes are not subject to shear buckling and are also eligible for the more liberal safety and resistance factors, v = 1.00 (LRFD) and v = 1.50 (ASD). This is presented in Example G.1 for a W-shape. A channel shear strength design is presented in Example G.2. A built-up girder with a thin web and transverse stiffeners is presented in Example G.8. G3. SINGLE ANGLES AND TEES A single angle example is illustrated in Example G.3.

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G4. RECTANGULAR HSS, BOX SECTIONS, AND OTHER SINGLY AND DOUBLY SYMMETRIC MEMBERS The shear height for HSS, h, is taken as the clear distance between the flanges less the inside corner radius on each side. If the corner radii are unknown, h shall be taken as the corresponding outside dimension minus 3 times the design thickness. A rectangular HSS example is provided in Example G.4. G5. ROUND HSS For all round HSS of ordinary length listed in the AISC Manual, Fcr can be taken as 0.6Fy in AISC Specification Equation G5-1. A round HSS example is illustrated in Example G.5. G6. WEAK AXIS SHEAR IN DOUBLY SYMMETRIC AND SINGLY SYMMETRIC SHAPES For examples of weak axis shear, see Example G.6 and Example G.7. G7. BEAMS AND GIRDERS WITH WEB OPENINGS For a beam and girder with web openings example, see AISC Design Guide 2, Design of Steel and Composite Beams with Web Openings (Darwin, 1990).

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EXAMPLE G.1A W-SHAPE IN STRONG AXIS SHEAR Given: Using AISC Manual tables, determine the available shear strength and adequacy of an ASTM A992 W2462 with end shears of 48 kips from dead load and 145 kips from live load. Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From Chapter 2 of ASCE/SEI 7, the required shear strength is: LRFD Vu  1.2  48 kips   1.6 145 kips 

ASD

Va  48 kips  145 kips  193 kips

 290 kips

From AISC Manual Table 3-2, the available shear strength is: LRFD

vVn  306 kips  290 kips o.k.

ASD Vn  204 kips  193 kips o.k. v

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EXAMPLE G.1B W-SHAPE IN STRONG AXIS SHEAR Given: The available shear strength of the W-shape in Example G.1A was easily determined using tabulated values in the AISC Manual. This example demonstrates the calculation of the available strength by directly applying the provisions of the AISC Specification. Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W2462 d = 23.7 in. tw = 0.430 in.

Nominal Shear Strength Except for very few sections, which are listed in the User Note, AISC Specification Section G2.1(a) is applicable to the I-shaped beams published in the AISC Manual for Fy  50 ksi. The W-shape sections that do not meet the criteria of AISC Specification Section G2.1(a) are indicated with footnote “v” in Tables 1-1, 3-2 and 6-2. Cv1 = 1.0

(Spec. Eq. G2-2)

From AISC Specification Section G2.1, area of the web, Aw, is determined as follows: Aw  dtw   23.7 in. 0.430 in.  10.2 in.2

From AISC Specification Section G2.1, the nominal shear strength is: Vn  0.6 Fy AwCv1



(Spec. Eq. G2-1)

 0.6  50 ksi  10.2 in.

2

 1.0 

 306 kips

Available Shear Strength From AISC Specification Section G2.1, the available shear strength is: LRFD v  1.00   vVn  1.00  306 kips   306 kips

ASD

v  1.50

Vn 306 kips  v 1.50  204 kips

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EXAMPLE G.2A CHANNEL IN STRONG AXIS SHEAR Given: Using AISC Manual tables, verify the available shear strength and adequacy of an ASTM A36 C1533.9 channel with end shears of 17.5 kips from dead load and 52.5 kips from live load.

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From Chapter 2 of ASCE/SEI 7, the required shear strength is: LRFD Vu  1.2 17.5 kips   1.6  52.5 kips 

ASD Va  17.5 kips  52.5 kips

 70.0 kips

 105 kips

From AISC Manual Table 3-8, the available shear strength is: LRFD vVn  117 kips  105 kips o.k.

ASD Vn  77.6 kips  70.0 kips o.k. v

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EXAMPLE G.2B CHANNEL IN STRONG AXIS SHEAR Given: The available shear strength of the channel in Example G.2A was easily determined using tabulated values in the AISC Manual. This example demonstrates the calculation of the available strength by directly applying the provisions of the AISC Specification.

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-5, the geometric properties are as follows: C1533.9

d = 15.0 in. tw = 0.400 in. Nominal Shear Strength All ASTM A36 channels listed in the AISC Manual have h tw  1.10 kv E / Fy ; therefore, Cv1 = 1.0

(Spec. Eq. G2-3)

From AISC Specification Section G2.1, the area of the web, Aw, is determined as follows: Aw  dtw  15.0 in. 0.400 in.  6.00 in.2

From AISC Specification Section G2.1, the nominal shear strength is: Vn  0.6 Fy AwCv1



(Spec. Eq. G2-1)



 0.6  36 ksi  6.00 in.2 1.0   130 kips

Available Shear Strength

Because AISC Specification Section G2.1(a) does not apply for channels, the values of v = 1.00 (LRFD) and v50 (ASD) may not be used. Instead v = 0.90 (LRFD) and v = 1.67 (ASD) from AISC Specification Section G1(a) must be used.

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LRFD v  0.90   vVn  0.90 130 kips   117 kips

ASD

v  1.67

Vn 130 kips  v 1.67  77.8 kips

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EXAMPLE G.3

ANGLE IN SHEAR

Given: Determine the available shear strength and adequacy of an ASTM A36 L534 (long leg vertical) with end shears of 3.5 kips from dead load and 10.5 kips from live load.

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-7, the geometric properties are as follows: L534

b = 5.00 in. t = 4 in. From Chapter 2 of ASCE/SEI 7, the required shear strength is: LRFD Vu  1.2  3.5 kips   1.6 10.5 kips 

ASD

Va  3.5 kips  10.5 kips  14.0 kips

 21.0 kips

Nominal Shear Strength Note: There are no tables in the AISC Manual for angles in shear, but the nominal shear strength can be calculated according to AISC Specification Section G3, as follows: From AISC Specification Section G3: kv = 1.2 Determine Cv2 from AISC Specification Section G2.2. h b  tw t 5.00 in.  4 in.  20.0

1.10

1.2  29, 000 ksi  kv E  1.10 Fy 36 ksi  34.2  20.0; therefore, use AISC Specification Equation G2-9

Cv2 = 1.0

(Spec. Eq. G2-9)

From AISC Specification Section G3, the nominal shear strength is:

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Vn  0.6 Fy btCv 2

(Spec. Eq. G3-1)

 0.6  36 ksi  5.00 in.4 in.1.0   27.0 kips

Available Shear Strength From AISC Specification Section G1, the available shear strength is: LRFD v  0.90   vVn  0.90  27.0 kips   24.3 kips  21.0 kips o.k.

ASD

v  1.67

Vn 27.0 kips  v 1.67  16.2 kips  14.0 kips o.k.

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EXAMPLE G.4

RECTANGULAR HSS IN SHEAR

Given: Determine the available shear strength by directly applying the provisions of the AISC Specification for an ASTM A500 Grade C HSS64a (long leg vertical) beam with end shears of 11 kips from dead load and 33 kips from live load. Note: There are tables in Part IV of this document that provide the shear strength of square and rectangular HSS shapes.

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C, rectangular Fy = 50 ksi Fu = 62 ksi From AISC Manual Table 1-11, the geometric properties are as follows: HSS64a

H = 6.00 in. B = 4.00 in. t = 0.349 in. From Chapter 2 of ASCE/SEI 7, the required shear strength is: LRFD Vu  1.2 11 kips   1.6  33 kips   66.0 kips

ASD

Va  11 kips  33 kips  44.0 kips

Nominal Shear Strength

The nominal shear strength can be determined from AISC Specification Section G4 as follows: The web shear buckling strength coefficient, Cv2, is found using AISC Specification Section G2.2 with h/tw = h/t and kv = 5. From AISC Specification Section G4, if the exact radius is unknown, h shall be taken as the corresponding outside dimension minus three times the design thickness.

h  H  3t  6.00 in.  3  0.349 in.  4.95 in. h 4.95 in.  t 0.349 in.  14.2

1.10

5  29, 000 ksi  kv E  1.10 Fy 50 ksi  59.2  14.2; therefore use AISC Specification Equation G2-9

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Cv2 = 1.0

(Spec. Eq. G2-9)

Note: Most standard HSS sections listed in the AISC Manual have Cv2 = 1.0 at Fy  50 ksi. Calculate Aw. Aw  2ht  2  4.95 in. 0.349 in.  3.46 in.2

Calculate Vn. Vn  0.6 Fy Aw Cv 2



(Spec. Eq. G4-1)



 0.6  50 ksi  3.46 in.2 1.0   104 kips

Available Shear Strength From AISC Specification Section G1, the available shear strength is: LRFD v  0.90   vVn  0.90 104 kips   93.6 kips  66.0 kips o.k.

ASD

v  1.67 Vn 104 kips  v 1.67  62.3 kips  44.0 kips o.k.

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EXAMPLE G.5

ROUND HSS IN SHEAR

Given:

Determine the available shear strength by directly applying the provisions of the AISC Specification for an ASTM A500 Grade C round HSS16.0000.375 beam spanning 32 ft with end shears of 30 kips from uniform dead load and 90 kips from uniform live load. Note: There are tables in Part IV of this document that provide the shear strength of round HSS shapes. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C, round HSS Fy = 46 ksi Fu = 62 ksi From AISC Manual Table 1-13, the geometric properties are as follows: HSS16.0000.375  A = 17.2 in.2 D/t = 45.8

From Chapter 2 of ASCE/SEI 7, the required shear strength is: LRFD Vu  1.2  30 kips   1.6  90 kips 

ASD

Va  30 kips  90 kips  120 kips

 180 kips Nominal Shear Strength

The nominal strength can be determined from AISC Specification Section G5, as follows: Using AISC Specification Section G5, calculate Fcr as the larger of: Fcr 

1.60 E

(Spec. Eq. G5-2a)

5

Lv  D  4   D t 

and Fcr 

0.78E 3 D 2

, but not to exceed 0.6 Fy

(Spec. Eq. G5-2b)

    t 

where Lv is taken as the distance from maximum shear force to zero; in this example, half the span.

Lv  0.5  32 ft 12 in./ft   192 in.

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Fcr 



1.60 E

(Spec. Eq. G5-2a)

5

Lv  D  4   D t  1.60  29, 000 ksi 

192 in.  45.85/4 16.0 in.  112 ksi Fcr 



0.78 E

(Spec. Eq. G5-2b)

3 D 2

    t  0.78  29, 000 ksi 

 45.83/ 2

 73.0 ksi

The maximum value of Fcr permitted is, Fcr  0.6 Fy  0.6  46 ksi   27.6 ksi

controls

Note: AISC Specification Equations G5-2a and G5-2b will not normally control for the sections published in the AISC Manual except when high strength steel is used or the span is unusually long. Calculate Vn using AISC Specification Section G5. Vn = 

Fcr Ag

(Spec. Eq. G5-1)

2

 27.6 ksi  17.2 in.2  2

 237 kips Available Shear Strength From AISC Specification Section G1, the available shear strength is: LRFD v  0.90   vVn  0.90  237 kips   213 kips  180 kips o.k.

ASD

v  1.67 Vn 237 kips  v 1.67  142 kips  120 kips o.k.

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EXAMPLE G.6

DOUBLY SYMMETRIC SHAPE IN WEAK AXIS SHEAR

Given: Verify the available shear strength and adequacy of an ASTM A992 W2148 beam with end shears of 20.0 kips from dead load and 60.0 kips from live load in the weak direction.

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W2148 bf = 8.14 in. tf = 0.430 in.

From Chapter 2 of ASCE/SEI 7, the required shear strength is: ASD Va  20.0 kips  60.0 kips

LRFD Vu  1.2  20.0 kips   1.6  60.0 kips 

 80.0 kips

 120 kips Nominal Shear Strength

From AISC Specification Section G6, for weak axis shear, use AISC Specification Equation G6-1. Calculate Cv2 using AISC Specification Section G2.2 with h tw  b f 2t f and kv = 1.2. bf h  tw 2t f 

8.14 in. 2  0.430 in.

 9.47 1.10

1.2  29, 000 ksi  kv E  1.10 50 ksi Fy  29.0  9.47

Therefore, use AISC Specification Equation G2-9:

Cv 2  1.0 Note: From the User Note in AISC Specification Section G6, Cv2 = 1.0 for all ASTM A6 W-, S-, M- and HP-shapes when Fy < 70 ksi. Calculate Vn. (Multiply the flange area by two to account for both shear resisting elements.)

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Vn  0.6 Fy b f t f Cv 2  2 

(from Spec. Eq. G6-1)

 0.6  50 ksi  8.14 in. 0.430 in.1.0  2   210 kips

Available Shear Strength From AISC Specification Section G1, the available shear strength is: ASD

LRFD

v  0.90   vVn  0.90  210 kips   189 kips  120 kips o.k.

v  1.67 Vn 210 kips  v 1.67  126 kips  80.0 kips o.k.

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EXAMPLE G.7

SINGLY SYMMETRIC SHAPE IN WEAK AXIS SHEAR

Given:

Verify the available shear strength and adequacy of an ASTM A36 C920 channel with end shears of 5 kips from dead load and 15 kips from live load in the weak direction. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-5, the geometric properties are as follows: C920 bf = 2.65 in. tf = 0.413 in.

From Chapter 2 of ASCE/SEI 7, the required shear strength is: ASD

LRFD Vu  1.2  5 kips   1.6 15 kips 

Vu  5 kips  15 kips  20.0 kips

 30.0 kips Nominal Shear Strength

Note: There are no AISC Manual tables for weak-axis shear in channel sections, but the available strength can be determined from AISC Specification Section G6. Calculate Cv2 using AISC Specification Section G2.2 with h/tw = bf /tf and kv = 1.2. h bf  tw t f 2.65 in. 0.413 in.  6.42 

1.10

1.2  29, 000 ksi  kv E  1.10 36 ksi Fy  34.2  6.42

Therefore, use AISC Specification Equation G2-9:

Cv 2  1.0 Calculate Vn. (Multiply the flange area by two to account for both shear resisting elements.)

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Vn  0.6 Fy b f t f Cv 2  2 

(from Spec. Eq. G6-1)

 0.6  36 ksi  2.65 in. 0.413 in.1.0  2   47.3 kips

Available Shear Strength From AISC Specification Section G1, the available shear strength is: ASD

LRFD

v  0.90   vVn  0.90  47.3 kips   42.6 kips  30.0 kips o.k.

v  1.67 Vn 47.3 kips  v 1.67  28.3 kips  20.0 kips o.k.

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EXAMPLE G.8A BUILT-UP GIRDER WITH TRANSVERSE STIFFENERS Given:

Determine the available shear strength of a built-up I-shaped girder for the span and loading as shown in Figure G.8A. The girder is ASTM A36 material and 36 in. deep with 16-in. 1½-in. flanges and a c-in.-thick web. The compression flange is continuously braced. Determine if the member has sufficient available shear strength to support the end shear, without and with tension field action. Use transverse stiffeners, as required. Note: This built-up girder was purposely selected with a thin web in order to illustrate the design of transverse stiffeners. A more conventionally proportioned plate girder may have at least a ½-in.-thick web and slightly smaller flanges.

Fig. G.8A. Beam loading and bracing diagram. Solution:

From AISC Manual Table 2-5, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi The geometric properties are as follows: Built-up girder tw = c in. d = 36.0 in. bft = bfc = 16.0 in. tf = 12 in. h = 33.0 in. From Chapter 2 of ASCE/SEI 7, the required shear strength at the support is: LRFD wu  1.2 1.06 kip/ft   1.6  3.13 kip/ft   6.28 kip/ft

Vu  

wu L 2  6.28 kip/ft  56 ft  2

 176 kips

ASD wa  1.06 kip/ft  3.13 kip/ft

 4.19 kip/ft Va  

wa L 2  4.19 kip/ft  56 ft  2

 117 kips

Stiffener Requirement Check

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From AISC Specification Section G2.1: Aw  dtw   36.0 in. c in.  11.3 in.2

For webs without transverse stiffeners, kv = 5.34 from AISC Specification Section G2.1(b)(2)(i).

h 33.0 in.  tw c in.  106 kv E  1.10 Fy

1.10

 5.34  29,000 ksi  36 ksi

 72.1  106 Therefore, use AISC Specification Equation G2-4:

Cv1 

1.10 kv E Fy

(Spec. Eq. G2-4)

h tw

72.1 106  0.680 

Calculate Vn. Vn  0.6 Fy AwCv1



(Spec. Eq. G2-1)



 0.6  36 ksi  11.3 in.2  0.680   166 kips

From AISC Specification Section G1, the available shear strength without stiffeners is: LRFD

ASD

v  0.90   vVn  0.90 166 kips   149 kips  176 kips n.g.

v  1.67 

Therefore, stiffeners are required.

Vn 166 kips  v 1.67  99.4 kips  117 kips n.g.

Therefore, stiffeners are required.

AISC Manual Tables 3-16a and 3-16b can be used to select the stiffener spacing needed to develop the required stress in the web. Stiffener Spacing for End Panel

Tension field action is not permitted for end panels, therefore use AISC Manual Table 3-16a.

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LRFD Use Vu = vVn to determine the required stress in the web by dividing by the web area.

ASD Use Va = Vn /v to determine the required stress in the web by dividing by the web area.

vVn Vu  Aw Aw 176 kips  11.3 in.2  15.6 ksi

Vn V  a v Aw Aw 117 kips  11.3 in.2  10.4 ksi

Use Table 3-16a from the AISC Manual to select the required stiffener ratio a/h based on the h/tw ratio of the girder and the required stress. Interpolate and follow an available stress curve, vVn/Aw= 15.6 ksi for LRFD, Vn/vAw = 10.4 ksi for ASD, until it intersects the horizontal line for an h/tw value of 106. Project down from this intersection and approximate the value for a/h as 1.40 from the axis across the bottom. Because h = 33.0 in., stiffeners are required at (1.40)(33.0 in.) = 46.2 in. maximum. Conservatively, use a 42-in. spacing. Stiffener Spacing for the Second Panel

From AISC Specification Section G2.2, tension field action is allowed because the second panel is an interior web panel. However, a web panel aspect ratio, a/h, must not exceed three. The required shear strength at the start of the second panel, 42 in. from the end, is: LRFD Vu  176 kips   6.28 kip/ft  42.0 in.1 ft/12 in.  154 kips

ASD Va  117 kips   4.19 kip/ft  42.0 in.1 ft/12 in.  102 kips

From AISC Specification Section G1, the available shear strength without stiffeners is: LRFD

ASD

v  0.90 

v  1.67

From previous calculations, vVn  149 kips  154 kips n.g. 

From previous calculations, Vn  99.4 kips  102 kips n.g. v

Therefore, additional stiffeners are required.

Therefore, additional stiffeners are required.

Use Vu = vVn to determine the required stress in the web by dividing by the web area.

Use Va = Vn /v to determine the required stress in the web by dividing by the web area.

vVn Vu  Aw Aw 154 kips  11.3 in.2  13.6 ksi

Vn V  a v Aw Aw 102 kips  11.3 in.2  9.03 ksi

Table 3-16b from the AISC Manual, including tension field action, may be used to select the required stiffener ratio a/h based on the h/tw ratio of the girder and the required stress, provided that the limitations of 2Aw / (Afc + Aft) ≤ 2.5, h/bfc ≤ 6.0, and h/bft ≤ 6.0 are met.

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



2 11.3 in.2 2 Aw  A fc  A ft 16.0 in.12 in.  16.0 in.12 in.  0.471  2.5 o.k. h h  b fc b ft 33.0 in. 16.0 in.  2.06  6.0 

o.k.

The limitations have been met. Table 3-16b may be used. Interpolate and follow an available stress curve, vVn/Aw = 13.6 ksi for LRFD, Vn/vAw = 9.03 ksi for ASD, until it intersects the horizontal line for an h/tw value of 106. Because the available stress does not intersect the h/tw value of 106, the maximum value of 3.0 for a/h may be used. Because h = 33.0 in., an additional stiffener is required at (3.0)(33.0 in.) = 99.0 in. maximum from the previous one. Conservatively, 90.0 in. spacing may be used. Stiffener Spacing for the Third Panel

From AISC Specification Section G2.2, tension field action is allowed because the next panel is not an end panel. The required shear strength at the start of the third panel, 132 in. from the end is: LRFD Vu  176 kips   6.28 kip/ft 132 in.1 ft/12 in.  107 kips

ASD Va  117 kips   4.19 kip/ft 132 in.1 ft/12 in.  70.9 kips

From AISC Specification Section G1, the available shear strength without stiffeners is: ASD

LRFD v  0.90

v  1.67

From previous calculations,

vVn  149 kips  107 kips o.k.

From previous calculations, Vn  99.4 kips  70.9 kips o.k. v

Therefore, additional stiffeners are not required.

Therefore, additional stiffeners are not required.

The six tables in the AISC Manual, 3-16a, 3-16b, 3-16c, 3-17a, 3-17b and 3-17c, are useful because they permit a direct solution for the required stiffener spacing. Alternatively, you can select a stiffener spacing and check the resulting strength, although this process is likely to be iterative. In Example G.8B, the stiffener spacings used are taken from this example.

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EXAMPLE G.8B BUILT-UP GIRDER WITH TRANSVERSE STIFFENERS Given: Verify the available shear strength and adequacy of the stiffener spacings from Example G.8A, which were easily determined from the tabulated values of the AISC Manual, by directly applying the provisions of the AISC Specification. Stiffeners are spaced at 42 in. in the first panel and 90 in. in the second panel. Solution: From AISC Manual Table 2-5, the material properties are as follows: ASTM A36 Fy = 36 ksi Fu = 58 ksi From Example G.8A, the required shear strength at the support is: ASD

LRFD Vu  176 kips

Va  117 kips

Shear Strength of End Panel

The web plate bucking coefficient, kv, is determined from AISC Specification Equation G2-5.

h 33.0 in.  tw c in.  106 kv  5   5

5

(Spec. Eq. G2-5)

 a h 2 5

 42.0 in. / 33.0 in.2

 8.09 1.10

8.09  29, 000 ksi  kv E  1.10 36 ksi Fy  88.8  106

Therefore, use AISC Specification Equation G2-4.

Cv1 

1.10 kv E Fy

(Spec. Eq. G2-4)

h tw

88.8 106  0.838 

Calculate Vn. From Example G.8A:

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Aw = 11.3 in.2 Vn  0.6 Fy AwCv1

(Spec. Eq. G2-1)





 0.6  36 ksi  11.3 in.2  0.838   205 kips

From AISC Specification Section G1, the available shear strength for the end panel is: ASD

LRFD

v  1.67

v  0.90   vVn  0.90  205 kips   185 kips  176 kips o.k.



Vn 205 kips  v 1.67  123 kips  117 kips o.k.

Shear Strength of the Second Panel

From Example G.8A, the required shear strength at the start of the second panel is: ASD

LRFD Vu  154 kips

Va  102 kips

The web plate bucking coefficient, kv, is determined from AISC Specification Equation G2-5.

kv  5   5

5

(Spec. Eq. G2-5)

 a h 2 5

 90.0 in. / 33.0 in.2

 5.67 1.37

5.67  29, 000 ksi  kv E  1.37 36 ksi Fy  92.6  106

Therefore, use AISC Specification Equation G2-11 to calculate Cv2.

Cv 2 

1.51kv E

(Spec. Eq. G2-11)

 h tw 2 Fy 1.51 5.67  29, 000 ksi   106 2  36 ksi   0.614

The limitations of AISC Specification Section G2.2(b)(1) are checked as follows:

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



2 11.3 in.2 2 Aw  A fc  A ft 16.0 in.12 in.  16.0 in.12 in.  0.471  2.5 h h  b fc b ft 33.0 in. 16.0 in.  2.06  6.0 

Because 2Aw / (Afc + Aft) ≤ 2.5, h/bfc ≤ 6.0, and h/bft ≤ 6.0, use AISC Specification Equation G2-7 with a = 90.0 in..  1  Cv 2 Vn  0.6 Fy Aw Cv 2   2 1.15 1   a h  

   

1  0.614   0.6  36 ksi  11.3 in.2 0.614  2  90.0 in.    1.15 1     33.0 in.    178 kips





(Spec. Eq. G2-7)     

From AISC Specification Section G1, the available shear strength for the second panel is: ASD

LRFD v  0.90   vVn  0.90 178 kips   160 kips  154 kips o.k.

v  1.67

Vn 178 kips  v 1.67  107 kips  102 kips o.k.

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CHAPTER G DESIGN EXAMPLE REFERENCES Darwin, D. (1990), Steel and Composite Beams with Web Openings, Design Guide 2, AISC, Chicago, IL.

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Chapter H Design of Members for Combined Forces and Torsion For all interaction equations in AISC Specification Chapter H, the required forces and moments must include second-order effects, as required by Chapter C of the AISC Specification. ASD users of the 1989 AISC Specification are accustomed to using an interaction equation that includes a partial second-order amplification. Second-order effects are now addressed in the analysis and are not included in these interaction equations.

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EXAMPLE H.1A W-SHAPE SUBJECT TO COMBINED COMPRESSION AND BENDING ABOUT BOTH AXES (BRACED FRAME) Given: Using Table IV-5 (located in this document), determine if an ASTM A992 W1499 has sufficient available strength to support the axial forces and moments listed as follows, obtained from a second-order analysis that includes P- effects. The unbraced length is 14 ft and the member has pinned ends. LRFD

ASD

Pu  400 kips M ux  250 kip-ft M uy  80.0 kip-ft

Pa  267 kips M ax  167 kip-ft M ay  53.3 kip-ft

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi The effective length of the member is: Lcx  Lcy  KL  1.0 14 ft   14.0 ft

For Lc = 14 ft, the combined strength parameters from Table IV-5 are: LRFD

p

ASD

0.887

p=

103 kips

bx 

by 

1.38 10 kip-ft 2.85

by =

3

10 kip-ft

Check Pr/Pc limit for AISC Specification Equation H1-1a.

 0.887  =  3   400 kips   10 kips   0.355

103 kips

bx =

3

Pu = pPu c Pn

1.33

2.08 3

10 kip-ft 4.29 3

10 kip-ft

Check Pr/Pc limit for AISC Specification Equation H1-1a.

Pa = pPa Pn / c  1.33  =  3   267 kips   10 kips   0.355

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LRFD

ASD

Because pPu  0.2,

pPu  bx M ux  by M uy  1.0

Because pPa  0.2, (from Part IV, Eq. IV-8)

pPa  bx M ax  by M ay  1.0

(from Part IV, Eq. IV-8)

 1.38   0.355   3   250 kip-ft   10 kip-ft 

 2.08   0.355   3  167 kip-ft   10 kip-ft 

 2.85    3   80.0 kip-ft   1.0  10 kip-ft   0.928  1.0 o.k.

 4.29    3   53.3kip-ft   1.0  10 kip-ft   0.931  1.0 o.k.

Table IV-5 simplifies the calculation of AISC Specification Equations H1-1a and H1-1b. A direct application of these equations is shown in Example H.1B.

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EXAMPLE H.1B W-SHAPE SUBJECT TO COMBINED COMPRESSION AND BENDING MOMENT ABOUT BOTH AXES (BRACED FRAME) Given: Using AISC Manual tables to determine the available compressive and flexural strengths, determine if an ASTM A992 W1499 has sufficient available strength to support the axial forces and moments listed as follows, obtained from a second-order analysis that includes P-  effects. The unbraced length is 14 ft and the member has pinned ends. LRFD Pu  400 kips M ux  250 kip-ft M uy  80 kip-ft

ASD Pa  267 kips M ax  167 kip-ft M ay  53.3 kip-ft

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi The effective length of the member is: Lcx  Lcy  KL  1.0 14 ft   14.0 ft

For Lc = 14.0 ft, the available axial and flexural strengths from AISC Manual Table 6-2 are: LRFD Pc  c Pn  1,130 kips

M cx  b M nx  642 kip-ft

M cy  b M ny  311 kip-ft

Pu 400 kips  c Pn 1,130 kips  0.354

ASD P Pc  n c  750 kips M nx b  427 kip-ft

M cx 

M ny b  207 kip-ft

M cy 

Pa 267 kips  Pn / c 750 kips  0.356

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LRFD

ASD

P Because u  0.2, c Pn

M ry Pr 8 M +  rx + Pc M cy 9  M cx 

Pa  0.2, Because Pn / c

   1.0 

(Spec. Eq. H1-1a)

400 kips 8  250 kip-ft 80.0 kip-ft  +  +   1.0 1,130 kips 9  642 kip-ft 311 kip-ft 

 0.928  1.0

o.k.

M ry  Pr 8 M (Spec. Eq. H1-1a) +  rx +   1.0 Pc M cy  9  M cx 267 kips 8  167 kip-ft 53.3 kip-ft   +  +  750 kips 9  427 kip-ft 207 kip-ft   0.932  1.0

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EXAMPLE H.2 W-SHAPE SUBJECT TO COMBINED COMPRESSION AND BENDING MOMENT ABOUT BOTH AXES (BY AISC SPECIFICATION SECTION H2) Given:

Using AISC Specification Section H2, determine if an ASTM A992 W1499 has sufficient available strength to support the axial forces and moments listed as follows, obtained from a second-order analysis that includes P-  effects. The unbraced length is 14 ft and the member has pinned ends. This example is included primarily to illustrate the use of AISC Specification Section H2. LRFD

ASD

Pu  360 kips M ux  250 kip-ft M uy  80 kip-ft

Pa  240 kips M ax  167 kip-ft M ay  53.3 kip-ft

Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W1499 A = 29.1 in.2 Sx = 157 in.3 Sy = 55.2 in.3

The required flexural and axial stresses are: ASD

LRFD

f ra

P  u A 360 kips   29.1 in.2  12.4 ksi

f ra





f rbx  

M ux Sx

f rbx 

 250 kip-ft 12 in./ft  3

157 in.  19.1 ksi f rby  

P  a A 240 kips   29.1 in.2  8.25 ksi

M uy





 80 kip-ft 12 in./ft  3

55.2 in.  17.4 ksi

167 kip-ft 12 in./ft 

157 in.3  12.8 ksi f rby 

Sy

M ax Sx



M ay Sy

 53.3 kip-ft 12 in./ft 

55.2 in.3  11.6 ksi

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The effective length of the member is: Lcx  Lcy  KL  1.0 14 ft   14.0 ft

For Lc = 14.0 ft, calculate the available axial and flexural stresses using the available strengths from AISC Manual Table 6-2. ASD Fcr Fca  c P  n  c A 750 kips  29.1 in.2  25.8 ksi  M Fcbx  nx b S x  427 kip-ft 12 in./ft    157 in.3  32.6 ksi  M ny Fcby  b S y

LRFD

Fca  c Fcr  

c Pn A  1,130 kips

29.1 in.2  38.8 ksi  

Fcbx  

b M nx Sx

 642 kip-ft 12 in./ft 

157 in.3  49.1 ksi Fcby  



b M ny Sy

 311 kip-ft 12 in./ft  3

55.2 in.  67.6 ksi





 207 kip-ft 12 in./ft 

55.2 in.3  45.0 ksi



As shown in the LRFD calculation of Fcby in the preceding text, the available flexural stresses can exceed the yield stress in cases where the available strength is governed by yielding and the yielding strength is calculated using the plastic section modulus. Combined Stress Ratio From AISC Specification Section H2, check the combined stress ratios as follows: ASD

LRFD f rby f ra f + rbx +  1.0 Fca Fcbx Fcby

(from Spec. Eq. H2-1)

12.4 ksi 19.1 ksi 17.4 ksi + +  0.966  1.0 o.k. 38.8 ksi 49.1 ksi 67.6 ksi

f rby f ra f + rbx +  1.0 Fca Fcbx Fcby

(from Spec. Eq. H2-1)

8.25 ksi 12.8 ksi 11.6 ksi + +  0.970  1.0 25.8 ksi 32.6 ksi 45.0 ksi

o.k.

A comparison of these results with those from Example H.1B shows that AISC Specification Equation H1-1a will produce less conservative results than AISC Specification Equation H2-1 when its use is permitted.

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Note: This check is made at a point on the cross section (extreme fiber, in this example). The designer must therefore determine which point on the cross section is critical, or check multiple points if the critical point cannot be readily determined.

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EXAMPLE H.3

W-SHAPE SUBJECT TO COMBINED AXIAL TENSION AND FLEXURE

Given:

Select an ASTM A992 W-shape with a 14-in.-nominal-depth to carry forces of 29 kips from dead load and 87 kips from live load in axial tension, as well as the following moments due to uniformly distributed loads: M xD  32 kip-ft M xL  96 kip-ft M yD  11.3 kip-ft M yL  33.8 kip-ft

The unbraced length is 30 ft and the ends are pinned. Assume the connections are made with no holes. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From ASCE/SEI 7, Chapter 2, the required strengths are: ASD

LRFD Pu  1.2  29 kips   1.6  87 kips 

Pa  29 kips  87 kips  116 kips

 174 kips

M ax  32 kip-ft  96 kip-ft

M ux  1.2  32 kip-ft   1.6  96 kip-ft 

 128 kip-ft

 192 kip-ft M uy  1.2 11.3 kip-ft   1.6  33.8 kip-ft   67.6 kip-ft

M ay  11.3 kip-ft  33.8 kip-ft  45.1 kip-ft

Try a W1482. From AISC Manual Tables 1-1 and 3-2, the properties are as follows: W1482

Ag = 24.0 in.2 Sx = 123 in.3 Zx = 139 in.3 Sy = 29.3 in.3 Zy = 44.8 in.3 Iy = 148 in.4 Lp = 8.76 ft Lr = 33.2 ft Nominal Tensile Strength From AISC Specification Section D2(a), the nominal tensile strength due to tensile yielding in the gross section is:

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Pn  Fy Ag

(Spec. Eq. D2-1)



  50 ksi  24.0 in.2



 1, 200 kips

Note that for a member with holes, the rupture strength of the member would also have to be computed using AISC Specification Equation D2-2. Nominal Flexural Strength for Bending About the Major Axis Yielding From AISC Specification Section F2.1, the nominal flexural strength due to yielding (plastic moment) is: M nx  M p  Fy Z x

(Spec. Eq. F2-1)



  50 ksi  139 in.

3



 6,950 kip-in.

Lateral-Torsional Buckling From AISC Specification Section F2.2, the nominal flexural strength due to lateral-torsional buckling is determined as follows: Because Lp < Lb M Lr, i.e., 8.76 ft < 30 ft < 33.2 ft, AISC Specification Equation F2-2 applies. Lateral-Torsional Buckling Modification Factor, Cb From AISC Manual Table 3-1, Cb = 1.14, without considering the beneficial effects of the tension force. However, per AISC Specification Section H1.2, Cb may be modified because the column is in axial tension concurrently with flexure. Pey  

2 EI y Lb 2



2  29, 000 ksi  148 in.4

 30 ft 12.0 in./ft    327 kips



2

LRFD

1

1.0 174 kips  Pu  1 327 kips Pey  1.24

ASD

1

1.6 116 kips  Pa  1 327 kips Pey  1.25

Cb  1.24 1.14   1.41

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  Lb  Lp  M n  Cb  M p   M p  0.7 Fy S x     M p   Lr  Lp 

(Spec. Eq. F2-2)

  1.416,950 kip-in.  6,950 kip-in.  0.7  50 ksi  123 in.3    6,560 kip-in. or 547 kip-ft controls



30 ft  8.76 ft    33.2   6,950 kip-in. ft  8.76 ft 

Local Buckling Per AISC Manual Table 1-1, the cross section is compact at Fy = 50 ksi; therefore, the local buckling limit state does not apply. Nominal Flexural Strength for Bending About the Minor Axis and the Interaction of Flexure and Tension Because a W1482 has compact flanges, only the limit state of yielding applies for bending about the minor axis. M ny  M p  Fy Z y  1.6 Fy S y



  50 ksi  44.8 in.

3

(Spec. Eq. F6-1)

  1.6 50 ksi   29.3 in.  3

 2, 240 kip-in.  2,340 kip-in. =2,240 kip-in. or 187 kip-ft

Available Strength From AISC Specification Sections D2 and F1, the available strengths are: ASD

LRFD b  t  0.90 

 Pc  t Pn

 0.90 1, 200 kips   1, 080 kips

M cx  b M nx  0.90  547 kip-ft 

 b  t  1.67   P Pc  n t 1, 200 kips  1.67  719 kips

M nx b 547 kip-ft = 1.67  328 kip-ft

M cx 

 492 kip-ft

 M cy  b M ny

 0.90 187 kip-ft   168 kip-ft

 M ny b 187 kip-ft  1.67  112 kip-ft

M cy 

Interaction of Tension and Flexure

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Check limit for AISC Specification Equation H1-1a. ASD

LRFD Pr P  u Pc t Pn 174 kips  1, 080 kips  0.161  0.2 Because

Pr Pa  Pc Pn / t 116 kips  719 kips  0.161  0.2

Pr  0.2, Pc

Because

Pr  M rx M ry  (Spec. Eq. H1-1b)     1.0 2 Pc  M cx M cy  174 kips 192 kip-ft 67.6 kip-ft     1.0 2 1, 080 kips  492 kip-ft 168 kip-ft

 0.873  1.0

o.k.

Pr  0.2, Pc

Pr  M rx M ry  (Spec. Eq. H1-1b)     1.0 2 Pc  M cx M cy  116 kips 128 kip-ft 45.1 kip-ft     1.0 2  719 kips  328 kip-ft 112 kip-ft

 0.874  1.0

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EXAMPLE H.4

W-SHAPE SUBJECT TO COMBINED AXIAL COMPRESSION AND FLEXURE

Given:

Select an ASTM A992 W-shape with a 10-in.-nominal-depth to carry axial compression forces of 5 kips from dead load and 15 kips from live load. The unbraced length is 14 ft and the ends are pinned. The member also has the following required moment strengths due to uniformly distributed loads, not including second-order effects: M xD  15 kip-ft M xL  45 kip-ft M yD  2 kip-ft M yL  6 kip-ft

The member is not subject to sidesway (no lateral translation). Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From Chapter 2 of ASCE/SEI 7, the required strength (not considering second-order effects) is: LRFD Pu  1.2  5 kips   1.6 15 kips   30.0 kips M ux  1.2 15 kip-ft   1.6  45 kip-ft   90.0 kip-ft M uy  1.2  2 kip-ft   1.6  6 kip-ft   12.0 kip-ft

ASD Pa  5 kips  15 kips  20.0 kips M ax  15 kip-ft  45 kip-ft  60.0 kip-ft M ay  2 kip-ft  6 kip-ft  8.00 kip-ft

Try a W1033. From AISC Manual Tables 1-1 and 3-2, the properties are as follows: W1033

A = 9.71 in.2 Sx = 35.0 in.3 Zx = 38.8 in.3 Ix = 171 in.4 rx = 4.19 in. Sy = 9.20 in.3 Zy = 14.0 in.3 Iy = 36.6 in.4 ry = 1.94 in. Lp = 6.85 ft Lr = 21.8 ft

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Available Axial Strength From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0. Because Lc = KLx = KLy = 14.0 ft and rx > ry, the y-y axis will govern. From AISC Manual Table 6-2, the available axial strength is: LRFD

ASD

Pc  c Pn  253 kips

P Pc  n c  168 kips

Required Flexural Strength (including second-order amplification) Use the approximate method of second-order analysis procedure from AISC Specification Appendix 8. Because the member is not subject to sidesway, only P- amplifiers need to be added. Cm 1 1  Pr / Pe1

B1 

(Spec. Eq. A-8-3)

where Cm is conservatively taken per AISC Specification A-8.2.1(b): Cm = 1.0 The x-x axis flexural magnifier is:

Pe1x 



2 EI x

(from Spec. Eq. A-8-5)

 Lc1x 2 2  29, 000 ksi  171 in.4 

14 ft 12 in./ft    1,730 kips

  1.0 

2

LRFD

  1.6 

Cm  1.0 1  Pr Pe1x 1.0   1.0 1  1.0  30 kips 1, 730 kips 

B1x 

 1.02

ASD

Cm  1.0 1  Pr Pe1x 1.0   1.0  1  1.6  20 kips 1,730 kips 

B1x 

M ux  1.02  90 kip-ft 

 1.02  M ax  1.02  60 kip-ft 

 91.8 kip-ft

 61.2 kip-ft

The y-y axis flexural magnifier is:

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Pe1 y 



2 EI y

 Lc1y 

(modified Spec. Eq. A-8-5)

2



2  29, 000 ksi  36.6 in.4

14 ft 12 in./ft    371 kips

LRFD

  1.0  B1y  



2

  1.6 

Cm  1.0 1  Pr Pe1y

B1y 

1.0  1.0 1  1.0  30 kips / 371 kips 



 1.09

ASD

Cm  1.0 1  Pr Pe1y 1.0  1.0  1  1.6  20 kips / 371kips 

 1.09  M ay  1.09  8 kip-ft 

M uy  1.09 12 kip-ft   13.1 kip-ft

 8.72 kip-ft

Nominal Flexural Strength about the Major Axis Yielding M nx  M p  Fy Z x



(Spec. Eq. F2-1)

  50 ksi  38.8 in.

3



 1,940 kip-in.

Lateral-Torsional Buckling Because Lp < Lb < Lr, i.e., 6.85 ft < 14.0 ft < 21.8 ft, AISC Specification Equation F2-2 applies. From AISC Manual Table 3-1, Cb = 1.14

  Lb  Lp  M nx  Cb  M p   M p  0.7 Fy S x     M p  Lr  Lp  

(Spec. Eq. F2-2)

  14 ft  6.85 ft    1.14 1,940 kip-in.  1,940 kip-in.  0.7  50 ksi  35.0 in.3       21.8 ft  6.85 ft     1,820 kip-in.  1,940 kip-in.  1,820 kip-in. or 152 kip-ft controls





Local Buckling Per AISC Manual Table 1-1, the member is compact for Fy = 50 ksi, so the local buckling limit state does not apply. Nominal Flexural Strength about the Minor Axis

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Determine the nominal flexural strength for bending about the minor axis from AISC Specification Section F6. Because a W1033 has compact flanges, only the yielding limit state applies. From AISC Specification Section F6.1: M nx  M p  Fy Z x  1.6 Fy S y





(Spec. Eq. F6-1)



  50 ksi  14.0 in.3  1.6  50 ksi  9.20 in.3



 700 kip-in.  736 kip-in.  700 kip-in. or 58.3 kip-ft

From AISC Specification Section F1, the available flexural strength is: ASD

LRFD  b  1.67

b  0.90

M cx  b M nx

M nx b 152 kip-ft  1.67  91.0 kip-ft

M cx 

 0.90 152 kip-ft   137 kip-ft

 M cy  b M ny

M ny b 58.3 kip-ft  1.67  34.9 kip-ft

M cy 

 0.90  58.3 kip-ft   52.5 kip-ft

Check limit for AISC Specification Equations H1-1a and H1-1b. ASD

LRFD

Pr P  u Pc c Pn 30 kips  253 kips  0.119  0.2 Because

Pr Pa  Pc Pn / c 20 kips  168 kips  0.119  0.2

Pr  0.2, Pc

M M ry Pr +  rx + 2 Pc M cy  M cx

Because    1.0 

(Spec. Eq. H1-1b)

 91.8 kip-ft 30 kips 13.1 kip-ft  + +   1.0 2  253 kips   137 kip-ft 52.5 kip-ft   0.979  1.0 o.k. 

Pr  0.2, Pc

M M ry Pr +  rx + 2 Pc M cy  M cx

   1.0 

(Spec. Eq. H1-1b)

 61.2 kip-ft 20 kips 8.72 kip-ft  + +  2 168 kips   91.0 kip-ft 34.9 kip-ft   0.982  1.0 o.k. 

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EXAMPLE H.5A RECTANGULAR HSS TORSIONAL STRENGTH Given:

Determine the available torsional strength of an ASTM A500, Grade C, HSS644. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi From AISC Manual Table 1-11, the geometric properties are as follows: HSS644 t = 0.233 in. b/t = 14.2 h/t = 22.8 C = 10.1 in.3

The available torsional strength for rectangular HSS is stipulated in AISC Specification Section H3.1. The critical stress, Fcr, is determined from AISC Specification Section H3.1(b). Because h/t > b/t, h/t governs.

2.45

E 29,000 ksi  2.45 50 ksi Fy  59.0  22.8; therefore, use AISC Specification Equation H3-3 to determine Fcr

Fcr  0.6 Fy

(Spec. Eq. H3-3)

 0.6  50 ksi   30.0 ksi

The nominal torsional strength is: Tn  Fcr C



  30.0 ksi  10.1 in.

3

(Spec. Eq. H3-1)



 303 kip-in.

From AISC Specification Section H3.1, the available torsional strength is: ASD

LRFD T  0.90 

 T Tn  0.90  303 kip-in.  273 kip-in.

T  1.67   Tn 303 kip-in.  T 1.67  181 kip-in.

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Note: For more complete guidance on designing for torsion, see AISC Design Guide 9, Torsional Analysis of Structural Steel Members (Seaburg and Carter, 1997).

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EXAMPLE H.5B ROUND HSS TORSIONAL STRENGTH Given:

Determine the available torsional strength of an ASTM A500, Grade C, HSS5.0000.250 that is 14 ft long. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C Fy = 46 ksi Fu = 62 ksi From AISC Manual Table 1-13, the geometric properties are as follows: HSS5.0000.250

D t D/t C

= 5.00 in. = 0.233 in. = 21.5 = 7.95 in.3

The available torsional strength for round HSS is stipulated in AISC Specification Section H3.1.The critical stress, Fcr, is determined from AISC Specification Section H3.1(a). Calculate the critical stress as the larger of:

Fcr =

=

1.23E L D   D t 

(Spec. Eq. H3-2a)

54

1.23  29,000 ksi 

14 ft 12 in./ft 

5.00 in.  133 ksi

 21.55 4

and

Fcr =

=

0.60 E

(Spec. Eq. H3-2b)

32

D    t  0.60  29, 000 ksi 

 21.53 2

 175 ksi However, Fcr shall not exceed the following: 0.6 Fy  0.6  46 ksi   27.6 ksi

Therefore, Fcr  27.6 ksi.

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The nominal torsional strength is: Tn  Fcr C



  27.6 ksi  7.95 in.

3

(Spec. Eq. H3-1)



 219 kip-in.

From AISC Specification Section H3.1, the available torsional strength is: ASD

LRFD T  0.90 

 T Tn  0.90  219 kip-in.  197 kip-in.

T  1.67   Tn 219 kip-in.  T 1.67  131 kip-in.

Note: For more complete guidance on designing for torsion, see AISC Design Guide 9, Torsional Analysis of Structural Steel Members (Seaburg and Carter, 1997).

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EXAMPLE H.5C RECTANGULAR HSS COMBINED TORSIONAL AND FLEXURAL STRENGTH Given:

Verify the strength of an ASTM A500, Grade C, HSS644 loaded as shown. The beam is simply supported and is torsionally fixed at the ends. Bending is about the strong axis.

Fig. H.5C. Beam loading and bracing diagram. Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi From AISC Manual Table 1-11, the geometric properties are as follows: HSS644

t Ag b/t h/t ry Zx J

= 0.233 in. = 4.30 in.2 = 14.2 = 22.8 = 1.61 in. = 8.53 in.3 = 23.6 in.4

From ASCE/SEI 7, Chapter 2, the required strength is: LRFD wu  1.2  0.46 kip/ft   1.6 1.38 kip/ft   2.76 kip/ft

ASD wa  0.46 kip/ft  1.38 kip/ft  1.84 kip/ft

Calculate the maximum shear (at the supports) using AISC Manual Table 3-23, Case 1. ASD

LRFD Vr  Vu w L  u 2  2.76 kip/ft  8 ft   2  11.0 kips

Vr  Va w L  a 2 1.84 kip/ft 8 ft   2  7.36 kips

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Calculate the maximum torsion (at the supports). LRFD Tr  Tu w Le  u 2 2.76 kip/ft  8 ft  6 in.   2  66.2 kip-in.

ASD Tr  Ta w Le  a 2 1.84 kip/ft  8 ft  6 in.   2  44.2 kip-in.

Available Shear Strength Determine the available shear strength from AISC Specification Section G4. Using the provisions given in AISC Specification Section B4.1b(d), determine the web depth, d, as follows: h  6.00 in.  3  0.233 in.  5.30 in.

From AISC Specification Section G4: Aw  2ht  2  5.30 in. 0.233 in.  2.47 in.2 kv  5

The web shear buckling coefficient is determined from AISC Specification Section G2.2. 1.10

5  29, 000 ksi  kv E = 1.10 50 ksi Fy  59.2  22.8; therefore use AISC Specification Section G2.2(b)(i)

Cv 2  1.0

(Spec. Eq. G2-9)

The nominal shear strength from AISC Specification Section G4 is: Vn  0.6 Fy AwC2

(Spec. Eq. G4-1)





 0.6  50 ksi  2.47 in.2 1.0   74.1 kips

From AISC Specification Section G1, the available shear strength is:

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LRFD

ASD  v  1.67

 v  0.90

Vc  vVn

Vn v 74.1 kips  1.67  44.4 kips

Vc 

 0.90  74.1 kips   66.7 kips



Available Flexural Strength The available flexural strength is determined from AISC Specification Section F7 for rectangular HSS. For the limit state of flexural yielding, the nominal flexural strength is:

Mn  M p

(Spec. Eq. F7-1)

 Fy Z x



  50 ksi  8.53 in.3



 427 kip-in. Determine if the limit state of flange local buckling applies as follows: b t  14.2



Determine the flange compact slenderness limit from AISC Specification Table B4.1b, Case 17.  p  1.12 = 1.12

E Fy 29, 000 ksi 50 ksi

 27.0    p ; therefore, the flange is compact and the flange local buckling limit state does not apply

Determine if the limit state of web local buckling applies as follows: h t  22.8



Determine the web compact slenderness limit from AISC Specification Table B4.1b, Case 19.

 p  2.42  2.42

E Fy 29, 000 ksi 50 ksi

 58.3

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   p ; therefore, the web is compact and the web local buckling limit state does not apply

Determine if lateral-torsional buckling applies as follows: L p  0.13Ery

JAg

(Spec. Eq. F7-12)

Mp

 23.6 in.  4.30 in.  4

 0.13  29, 000 ksi 1.61 in.

2

427 kip-in.

 143 in. or 11.9 ft

Since Lb = 8 ft < Lp = 11.9 ft, lateral-torsional buckling is not applicable and Mn = 427 kip-in., controlled by the flexural yielding limit state. From AISC Specification Section F1, the available flexural strength is: LRFD

ASD  b  1.67   M Mc  n b 427 kip-in.  1.67  256 kip-in.

b  0.90

M c  b M n  0.90  427 kip-in.   384 kip-in.

From Example H.5A, the available torsional strength is: LRFD

ASD

Tc  T Tn

T Tc  n T  181 kip-in.

 273 kip-in.

Using AISC Specification Section H3.2, check combined strength at several locations where Tr > 0.2Tc. First check at the supports, which is the point of maximum shear and torsion: LRFD

ASD

Tr 66.2 kip-in. = Tc 273 kip-in.  0.242  0.2

Tr 44.2 kip-in. = Tc 181 kip-in.  0.244  0.2

Therefore, use AISC Specification Equation H3-6:

Therefore, use AISC Specification Equation H3-6:

 Pr M r P M c  c

2

  Vr Tr     V  T   1.0 c    c

(Spec Eq. H3-6)

 11.0 kips 66.2 kip-in.    0  0      66.7 kips 273 kip-in.   0.166  1.0

o.k.

2

 Pr M r P M c  c

2

  Vr Tr     V  T   1.0 c    c

(Spec Eq. H3-6)

 7.36 kips 44.2 kip-in.    0  0   +  181 kip-in.   44.4 kips  0.168  1.0

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Check the combined strength near the location where Tr = 0.2Tc. This is the location with the largest bending moment required to be considered in the interaction. Calculate the shear and moment at this location, x. LRFD

ASD

Tr  0.20 Tc

Tr  0.20 Tc

Therefore at x:

Therefore at x:

Tr  0.20  273 kip-in.

Tr  0.20 181 kip-in.

 54.6 kip-in.

x

Tr

 36.2 kip-in.

at support   Tr at x 

x

wu e 66.2 kip-in.  54.6 kip-in.   2.76 kip/ft  6 in.

 0.725 ft

Vr  11.0 kips   0.700 ft  2.76 kip/ft 

Vr  7.36 kips   0.725 ft 1.84 kips/ft 

 9.07 kips

 6.03 kips

wu x l  x  2  2.76 kip/ft  0.700 ft 

Mr 

8 ft  0.700 ft  2  7.05 kip-ft or 84.6 kip-in. 

at support   Tr at x 

wa e 44.2 kip-in.  36.2 kip-in.  1.84 kip/ft  6 in.

 0.700 ft

Mr 

Tr

2

 Pr M r   Vr Tr          1.0  Pc M c   Vc Tc 

wa x l  x  2 1.84 kip/ft  0.725 ft 

8 ft  0.725 ft  2  4.85 kip-ft or 58.2 kip-in. 

2

(Spec Eq. H3-6)

  84.6 kip-in.   9.07 kips  0   0.20   384 kip-in.   66.7 kips    0.333  1.0 o.k.

2

 Pr M r   Vr Tr          1.0  Pc M c   Vc Tc 

(Spec Eq. H3-6)

  58.2 kip-in.   6.03 kips  0   0.20  + 256 kip-in.   44.4 kips    0.340  1.0 o.k.

2

Note: The remainder of the beam, where Tr M 0.2Tc, must also be checked to determine if the strength without torsion controls over the interaction with torsion.

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EXAMPLE H.6

W-SHAPE TORSIONAL STRENGTH

Given: As shown in Figure H.6-1, an ASTM A992 W1049 spans 15 ft and supports concentrated loads at midspan that act at a 6-in. eccentricity with respect to the shear center. Determine the stresses on the cross section, the adequacy of the section to support the loads, and the maximum rotation.

Fig. H.6-1. Beam loading diagram. The end conditions are assumed to be flexurally pinned and unrestrained for warping torsion. The eccentric load can be resolved into a torsional moment and a load applied through the shear center. A similar design example appears in AISC Design Guide 9, Torsional Analysis of Structural Steel Members (Seaburg and Carter, 1997).

Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: W1049 tw = 0.340 in. tf = 0.560 in. Ix = 272 in.4 Sx = 54.6 in.3 Zx = 60.4 in.3 J = 1.39 in.4 Cw = 2,070 in.6

From the AISC Shapes Database, the additional torsional properties are as follows: W1049 Sw1 = 33.0 in.4 Wno = 23.6 in.2 Qf = 12.8 in.3 Qw = 29.8 in.3

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From AISC Design Guide 9, the torsional property, a, is calculated as follows: a 

ECw GJ

(Design Guide 9, Eq. 3.6)

 29, 000 ksi   2,070 in.6  11, 200 ksi  1.39 in.4 

 62.1 in. From ASCE/SEI 7, Chapter 2, and AISC Manual Table 3-23, Case 7, the required strengths are: ASD

LRFD Pu  1.2  2.5 kips   1.6  7.5 kips 

Pa  2.5 kips  7.5 kips  10.0 kips

 15.0 kips

Pa 2 10.0 kips  2  5.00 kips

Pu 2 15.0 kips  2  7.50 kips

Va 

Vu 

Mu  

Pu L 4 15.0 kips 15 ft 12 in./ft 

Ma  

4

Pa L 4 10.0 kips 15 ft 12 in./ft  4

 450 kip-in.

 675 kip-in.

Ta  Pa e

Tu  Pu e  15.0 kips  6 in.

 10.0 kips  6 in.

 90.0 kip-in.

 60.0 kip-in.

Normal and Shear Stresses from Flexure The normal and shear stresses from flexure are determined from AISC Design Guide 9, as follows:

ub

LRFD Mu (from Design Guide 9, Eq. 4.5)  Sx 675 kip-in.  54.6 in.3  12.4 ksi (compression at top, tension at bottom)

ub web = =

Vu Qw I x tw

(from Design Guide 9, Eq. 4.6)

 7.50 kips   29.8 in.3   272 in.4   0.340 in.

 2.42 ksi

ASD Ma (from Design Guide 9, Eq. 4.5)  ab = Sx 450 kip-in.  54.6 in.3  8.24 ksi (compression at top, tension at bottom)  ab web =



Va Qw I x tw

(from Design Guide 9, Eq. 4.6)

 5.00 kips   29.8 in.3   272 in.4   0.340 in.

 1.61 ksi

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LRFD ub flange =

Vu Q f

ASD

(from Design Guide 9, Eq. 4.6)

I xt f

ab flange =

Va Q f I xt f

(from Design Guide 9, Eq. 4.6)

 7.50 kips  12.8 in.3  =  272 in.4   0.560 in.

 5.00 kips  12.8 in.3  =  272 in.4   0.560 in.

 0.630 ksi

 0.420 ksi

Torsional Stresses The following functions are taken from AISC Design Guide 9, Appendix B, Case 3, with  = 0.5 for the torsional load applied at midspan.

L 15 ft 12 in./ft   a 62.1 in.  2.90 Using the graphs in AISC Design Guide 9, Appendix B, select values for , ,  and . At midspan (z/l = 0.5):

Tr l GJ

For :

 GJ   1       +0.09  Tr   l 

Solve for:   +0.09

For :

 GJ     Tr

Therefore:   0

For :

 GJ     a  0.44  Tr 

Solve for:   0.44

For :

 GJ     Tr

Solve for:   0.50

 0 

 2  a  0.50 

Tr GJa Tr GJa 2

At the support (z/l = 0): For :

 GJ   Tr

1  l   0  

For :

 GJ       0.28  Tr 

Solve for:   0.28

For :

 GJ     Tr

Therefore:   0

For :

 GJ     Tr

 a  0   2  a  0.22 

Therefore:   0 Tr GJ

Solve for:   0.22

Tr GJa 2

In the preceding calculations, note that the applied torque is negative based on the sign convention used in the AISC Design Guide 9 graphs. Calculate Tr/GJ as follows:

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LRFD Tu 90.0 kip-in. = GJ 11, 200 ksi  1.39 in.4

ASD Ta 60.0 kip-in. = GJ 11, 200 ksi  1.39 in.4

=  5.78  10 3 rad/in.

=  3.85  10 3 rad/in.









Shear Stresses Due to Pure Torsion The shear stresses due to pure torsion are determined from AISC Design Guide 9 as follows: t  Gt 

(Design Guide 9, Eq. 4.1) ASD

LRFD At midspan:

At midspan:

  0; therefore ut  0

  0; therefore  at  0

At the support, for the web:

At the support, for the web:

 5.78 rad  ut  11, 200 ksi  0.340 in. 0.28    103 in.   6.16 ksi

 3.85 rad  at  11, 200 ksi  (0.340 in.)(0.28)    103 in.  =  4.11 ksi

At the support, for the flange:

At the support, for the flange:

 5.78 rad  ut  11, 200 ksi  0.560 in. 0.28    103 in.  =  10.2 ksi

 3.85 rad  at  11, 200 ksi  0.560 in. 0.28    103 in.  =  6.76 ksi

Shear Stresses Due to Warping The shear stresses due to warping are determined from AISC Design Guide 9 as follows: w 

 ES w1 tf

(Design Guide 9, Eq. 4.2a) LRFD

ASD

At midspan:

uw 

At midspan:

 29, 000 ksi   33.0 in.4   0.560 in.

0.50  5.78 rad     62.1 in.2 103 in.   





=  1.28 ksi

=  0.563 ksi

0.560 in.

0.50  3.85 rad     62.1 in.2 103 in.   





At the support:

 29, 000 ksi   33.0 in.4   0.560 in.

 29, 000 ksi   33.0 in.4  

=  0.853 ksi

At the support:

uw 

aw 

0.22  5.78 rad     62.1 in.2 103 in.   





aw 

 29, 000 ksi   33.0 in.4   0.560 in.

=  0.375 ksi

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

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Normal Stresses Due to Warping The normal stresses due to warping are determined from AISC Design Guide 9 as follows:  w  EWno 

(Design Guide 9, Eq. 4.3a) ASD

LRFD At midspan:

At midspan:

 0.44 5.78 rad    uw   29, 000 ksi  23.6 in.2  3   62.1 in. 10 in.    = 28.0 ksi

 0.44 3.85 rad    aw   29, 000 ksi  23.6 in.2  3   62.1 in. 10 in.    = 18.7 ksi

At the support:

At the support:

Because   0, uw  0.

Because   0,  aw  0.

















Combined Stresses The stresses are summarized in Tables H.6-1A and H.6-1B and shown in Figure H.6-2.

Table H.6-1A Summary of Stresses Due to Flexure and Torsion (LRFD), ksi Location

Normal Stress

 uw

ub

Flange Web

28.0 –

12.4 –

Flange Web Maximum

0 –

0 –

Shear Stress

f un

ut

Midspan 0 40.4 – 0 Support 0 10.2 – 6.16 40.4

uw

ub

f uv

1.28 –

0.630 2.42

1.91 ±2.42

0.563 –

0.630 2.42

11.4 8.58 11.4

Table H.6-1B Summary of Stresses Due to Flexure and Torsion (ASD), ksi Normal Stress

Location

 aw

ab

Flange Web

18.7 –

8.24 –

Flange Web Maximum

0 –

0 –

Shear Stress

f an

at

Midspan 0 26.9 – 0 Support 0 6.76 – 4.11 26.9

 aw

ab

f av

0.853 –

0.420 1.61

1.27 ±1.61

0.375 –

0.420 1.61

7.56 5.72 7.56

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(a) Normal stresses due to flexure and torsion at midspan—LRFD

(b) Normal stresses due to flexure and torsion at midspan—ASD

(c) Shear stresses due to flexure and torsion at support—LRFD

(d) Shear stresses due to flexure and torsion at support—ASD

Fig. H.6-2. Stresses due to flexure and torsion. LRFD The maximum normal stress due to flexure and torsion occurs at the edge of the flange at midspan and is equal to 40.4 ksi.

ASD The maximum normal stress due to flexure and torsion occurs at the edge of the flange at midspan and is equal to 26.9 ksi.

The maximum shear stress due to flexure and torsion occurs in the middle of the flange at the support and is equal to 11.4 ksi.

The maximum shear stress due to flexure and torsion occurs in the middle of the flange at the support and is equal to 7.56 ksi.

Available Torsional Strength The available torsional strength is the lowest value determined for the limit states of yielding under normal stress, shear yielding under shear stress, or buckling in accordance with AISC Specification Section H3.3. The nominal torsional strength due to the limit states of yielding under normal stress and shear yielding under shear stress are compared to the applicable buckling limit states. Buckling For the buckling limit state, lateral-torsional buckling and local buckling must be evaluated. The nominal torsional strength due to the limit state of lateral-torsional buckling is determined as follows.

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Cb = 1.32 from AISC Manual Table 3-1. Compute Fn for a W1049 using values from AISC Manual Table 3-10 with Lb = 15 ft and Cb = 1.0. LRFD

ASD

b  0.90   b M n  204 kips

b  1.67 Mn  136 kip-ft b

Fn  Fcr

(Spec. Eq. H3-9)

Fn  Fcr

(Spec. Eq. H3-9)

M   Cb  n   Sx 

M   Cb  n   Sx   204 kip-ft 12 in./ft     1.32  3  0.90 54.6 in.   65.8 ksi





   



1.67 136 kip-ft 12 in./ft     1.32  3  54.6 in.   65.9 ksi





   

The limit state of local buckling does not apply because a W1049 is compact in flexure per the user note in AISC Specification Section F2. Yielding Under Normal Stress The nominal torsional strength due to the limit state of yielding under normal stress is determined as follows:

Fn  Fy

(Spec. Eq. H3-7)

 50 ksi Therefore, the limit state of yielding under normal stress controls over buckling. The available torsional strength for yielding under normal stress is determined as follows, from AISC Specification Section H3: LRFD T  0.90 

 T Fn  0.90  50 ksi   45.0 ksi  40.4 ksi

o.k.

ASD T  1.67   Fn 50 ksi  T 1.67  29.9 ksi  26.9 ksi o.k.

Shear Yielding Under Shear Stress The nominal torsional strength due to the limit state of shear yielding under shear stress is: Fn  0.6 Fy

(Spec. Eq. H3-8)

 0.6  50 ksi   30.0 ksi

The limit state of shear yielding under shear stress controls over buckling. The available torsional strength for shear yielding under shear stress is determined as follows, from AISC Specification Section H3:

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LRFD

ASD

T  0.90 

T  1.67   Fn 30 ksi  T 1.67  18.0 ksi  7.56 ksi

 T Fn  0.90  30 ksi   27.0 ksi  11.4 ksi

o.k.

o.k.

Maximum Rotation at Service Load The maximum rotation occurs at midspan. The service load torque is: T  Pe    2.50 kips  7.50 kips  6 in.  60.0 kip-in.

As determined previously from AISC Design Guide 9, Appendix B, Case 3 with  = 0.5, the maximum rotation is: Tl GJ 0.09  60.0 kip-in.15 ft 12 in./ft 

  0.09 

11,200 ksi  1.39 in.4 

 0.0624 rad or  3.58

See AISC Design Guide 9, Torsional Analysis of Structural Steel Members, for additional guidance.

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CHAPTER H DESIGN EXAMPLE REFERENCES Seaburg, P.A. and Carter, C.J. (1997), Torsional Analysis of Structural Steel Members, Design Guide 9, AISC, Chicago, IL.

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Chapter I Design of Composite Members I1.

GENERAL PROVISIONS

Design, detailing, and material properties related to the concrete and steel reinforcing portions of composite members are governed by ACI 318 (ACI 318, 2014) as modified with composite-specific provisions by the AISC Specification. The available strength of composite sections may be calculated by one of four methods: the plastic stress distribution method, the strain-compatibility method, the elastic stress distribution method, or the effective stress-strain method. The composite design tables in Part IV of this document are based on the plastic stress distribution method. Filled composite sections are classified for local buckling according to the slenderness of the compression steel elements as illustrated in AISC Specification Tables I1.1a and I1.1b, and Examples I.4, I.6 and I.7. Local buckling effects do not need to be considered for encased composite members. Terminology used within the Examples for filled composite section geometry is illustrated in Figure I-1. I2.

AXIAL FORCE

The available compressive strength of a composite member is based on a summation of the strengths of all of the components of the column with reductions applied for member slenderness and local buckling effects where applicable. For tension members, the concrete tensile strength is ignored and only the strength of the steel member and properly connected reinforcing is permitted to be used in the calculation of available tensile strength. The available compressive strengths for filled composite sections are given in Part IV of this document and reflect the requirements given in AISC Specification Sections I1.4 and I2.2. The design of filled composite compression and tension members is presented in Examples I.4 and I.5, respectively. The design of encased composite compression and tension members is presented in Examples I.9 and I.10, respectively. There are no tables in the AISC Manual for the design of these members. Note that the AISC Specification stipulates that the available compressive strength need not be less than that specified for the bare steel member. I3.

FLEXURE

The design of typical composite beams with steel anchors is illustrated in Examples I.1 and I.2. AISC Manual Table 3-19 provides available flexural strengths for composite W-shape beams, Table 3-20 provides lower-bound moments of inertia for plastic composite sections, and Table 3-21 provides shear strengths of steel headed stud anchors utilized for composite action in composite beams. The design of filled composite members for flexure is illustrated within Examples I.6 and I.7, and the design of encased composite members for flexure is illustrated within Example I.11. I4.

SHEAR

For composite beams with formed steel deck, the available shear strength is based upon the properties of the steel section alone in accordance with AISC Specification Chapter G as illustrated in Examples I.1 and I.2.

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For filled and encased composite members, either the shear strength of the steel section alone, the steel section plus the reinforcing steel, or the reinforced concrete alone are permitted to be used in the calculation of available shear strength. The calculation of shear strength for filled composite members is illustrated within Examples I.6 and I.7 and for encased composite members within Example I.11. I5.

COMBINED FLEXURE AND AXIAL FORCE

Design for combined axial force and flexure may be accomplished using either the strain compatibility method or the plastic-distribution method. Several different procedures for employing the plastic-distribution method are outlined in the Commentary, and each of these procedures is demonstrated for filled composite members in Example I.6 and for encased composite members in Example I.11. Interaction calculations for noncompact and slender filled composite members are illustrated in Example I.7. To assist in developing the interaction curves illustrated within the design examples, a series of equations is provided in AISC Manual Part 6, Tables 6-3a, 6-3b, 6-4 and 6-5. These equations define selected points on the interaction curve, without consideration of slenderness effects. Specific cases are outlined and the applicability of the equations to a cross section that differs should be carefully considered. As an example, the equations in AISC Manual Table 6-3a are appropriate for the case of side bars located at the centerline, but not for other side bar locations. In contrast, these equations are appropriate for any amount of reinforcing at the extreme reinforcing bar location. In AISC Manual Table 6-3b the equations are appropriate only for the case of four reinforcing bars at the corners of the encased section. When design cases deviate from those presented the appropriate interaction equations can be derived from first principles. I6.

LOAD TRANSFER

The AISC Specification provides several requirements to ensure that the concrete and steel portions of the section act together. These requirements address both force allocation—how much of the applied loads are resisted by the steel versus the reinforced concrete; and force transfer mechanisms—how the force is transferred between the two materials. These requirements are illustrated in Example I.3 for filled composite members and Example I.8 for encased composite members. I7.

COMPOSITE DIAPHRAGMS AND COLLECTOR BEAMS

The Commentary provides guidance on design methodologies for both composite diaphragms and composite collector beams. I8.

STEEL ANCHORS

AISC Specification Section I8 addresses the strength of steel anchors in composite beams and in composite components. Examples I.1 and I.2 illustrates the design of composite beams with steel headed stud anchors. The application of steel anchors in composite component provisions have strict limitations as summarized in the User Note provided at the beginning of AISC Specification Section I8.3. These provisions do not apply to typical composite beam designs nor do they apply to hybrid construction where the steel and concrete do not resist loads together via composite action such as in embed plates. The most common application for these provisions is for the transfer of longitudinal shear within the load introduction length of composite columns as demonstrated in Example I.8. The application of these provisions to an isolated anchor within an applicable composite system is illustrated in Example I.12.

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Fig. I-1. Terminology used for filled members.

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EXAMPLE I.1 COMPOSITE BEAM DESIGN Given: A typical bay of a composite floor system is illustrated in Figure I.1-1. Select an appropriate ASTM A992 W-shaped beam and determine the required number of w-in.-diameter steel headed stud anchors. The beam will not be shored during construction.

Fig. I.1-1. Composite bay and beam section. To achieve a two-hour fire rating without the application of spray applied fire protection material to the composite deck, 42 in. of normal weight (145 lb/ft3) concrete will be placed above the top of the deck. The concrete has a specified compressive strength, f c = 4 ksi. Applied loads are given in the following: Dead Loads: Pre-composite: Slab = 75 lb/ft2 (in accordance with metal deck manufacturer’s data) Self-weight = 5 lb/ft2 (assumed uniform load to account for beam weight) Composite (applied after composite action has been achieved): Miscellaneous = 10 lb/ft2 (HVAC, ceiling, floor covering, etc.) Live Loads: Pre-composite: Construction = 25 lb/ft2 (temporary loads during concrete placement)

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Composite (applied after composite action has been achieved): Non-reducible = 100 lb/ft2 (assembly occupancy) Solution: From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi Applied Loads For slabs that are to be placed at a constant elevation, AISC Design Guide 3 (West and Fisher, 2003) recommends an additional 10% of the nominal slab weight be applied to account for concrete ponding due to deflections resulting from the wet weight of the concrete during placement. For the slab under consideration, this would result in an additional load of 8 lb/ft2; however, for this design the slab will be placed at a constant thickness, and thus, no additional weight for concrete ponding is required. For pre-composite construction live loading, 25 lb/ft2 will be applied in accordance with recommendations from Design Loads on Structures During Construction, ASCE/SEI 37 (ASCE, 2014), for a light duty operational class that includes concrete transport and placement by hose and finishing with hand tools. Composite Deck and Anchor Requirements Check composite deck and anchor requirements stipulated in AISC Specification Sections I1.3, I3.2c and I8. 3 ksi  f c  10 ksi (for normal weight concrete)

(Spec. Section I1.3)

1.

Concrete Strength: f c  4 ksi o.k.

2.

Rib height: hr  3 in. hr  3 in. o.k.

(Spec. Section I3.2c)

3.

Average rib width: wr  2 in. wr  6 in. (from deck manufacturer’s literature) o.k.

(Spec. Section I3.2c)

4.

Use steel headed stud anchors w in. or less in diameter.

(Spec. Section I8.1)

Use w-in.-diameter steel anchors per problem statement. o.k. 5.

Steel headed stud anchor diameter: d sa  2.5t f

(Spec. Section I8.1)

In accordance with AISC Specification Section I8.1, this limit only applies if steel headed stud anchors are not welded to the flange directly over the web. The w-in.-diameter anchors will be placed in pairs transverse to the web in some locations, thus this limit must be satisfied. Select a beam size with a minimum flange thickness of 0.300 in., as determined in the following:

d sa 2.5 w in.  2.5  0.300 in.

tf 

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6.

In accordance with AISC Specification I3.2c, steel headed stud anchors, after installation, shall extend not less than 12 in. above the top of the steel deck. A minimum anchor length of 42 in. is required to meet this requirement for 3 in. deep deck. From steel headed stud anchor manufacturer’s data, a standard stock length of 4d in. is selected. Using a a-in. length reduction to account for burn off during anchor installation through the deck yields a final installed length of 42 in.

7.

Minimum length of stud anchors  4d sa 42 in.  4  w in.  3.00 in. o.k.

8.

In accordance with AISC Specification Section I3.2c, there shall be at least 2 in. of specified concrete cover above the top of the headed stud anchors.

(Spec. Section I8.2)

As discussed in AISC Specification Commentary to Section I3.2c, it is advisable to provide greater than 2 in. minimum cover to assure anchors are not exposed in the final condition, particularly for intentionally cambered beams. 72 in.  42 in.  3.00 in.  2 in. o.k.

9.

In accordance with AISC Specification Section I3.2c, slab thickness above steel deck shall not be less than 2 in. 42 in.  2 in. o.k.

Design for Pre-Composite Condition

Construction (Pre-Composite) Loads The beam is uniformly loaded by its tributary width as follows:





wD  10 ft  75 lb/ft 2  5 lb/ft 2  1 kip 1,000 lb     0.800 kip/ft





wL  10 ft  25 lb/ft 2  1 kip 1,000 lb     0.250 kip/ft

Construction (Pre-Composite) Flexural Strength From ASCE/SEI 7, Chapter 2, the required flexural strength is: LRFD

ASD

wu  1.2  0.800 kip/ft   1.6  0.250 kip/ft   1.36 kip/ft w L2 Mu  u 8 

1.36 kip/ft  45 ft 2

8  344 kip-ft

wa  0.800 kip/ft  0.250 kip/ft  1.05 kip/ft

Ma  

wa L2 8

1.05 kip/ft  45 ft 2

8  266 kip-ft

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Beam Selection Assume that attachment of the deck perpendicular to the beam provides adequate bracing to the compression flange during construction, thus the beam can develop its full plastic moment capacity. The required plastic section modulus, Zx, is determined as follows, from AISC Specification Equation F2-1: LRFD

ASD

b  0.90 Z x, min  

b  1.67

Mu b Fy

Z x , min 

 344 kip-ft 12 in./ft  0.90  50 ksi 



b M a Fy 1.67  266 kip-ft 12 in./ft  50 ksi 3

3

 107 in.

 91.7 in.

From AISC Manual Table 3-2, select a W2150 with a Zx value of 110 in.3 Note that for the member size chosen, the self-weight on a pounds per square foot basis is 50 plf 10 ft  5.00 psf ; thus the initial self-weight assumption is adequate.

From AISC Manual Table 1-1, the geometric properties are as follows: W2150

A tf h/tw Ix

= 14.7 in.2 = 0.535 in. = 49.4 = 984 in.4

Pre-Composite Deflections AISC Design Guide 3 (West and Fisher, 2003) recommends deflections due to concrete plus self-weight not exceed the minimum of L/360 or 1.0 in. From AISC Manual Table 3-23, Case 1:  nc 

5wD L4 384 EI

Substituting for the moment of inertia of the non-composite section, I  984 in.4 , yields a dead load deflection of:

 nc 

5  0.800 kip/ft 1 ft/12 in.  45 ft 12 in./ft  



384  29, 000 ksi  984 in.4

4



 2.59 in.  L / 208  L / 360

n.g.

Pre-composite deflections exceed the recommended limit. One possible solution is to increase the member size. A second solution is to induce camber into the member. For this example, the second solution is selected, and the beam will be cambered to reduce the net pre-composite deflections.

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Reducing the estimated simple span deflections to 80% of the calculated value to reflect the partial restraint of the end connections as recommended in AISC Design Guide 3 yields a camber of: Camber = 0.8  2.59 in.  2.07 in.

Rounding down to the nearest 4-in. increment yields a specified camber of 2 in. Select a W2150 with 2 in. of camber. Design for Composite Condition

Required Flexural Strength Using tributary area calculations, the total uniform loads (including pre-composite dead loads in addition to dead and live loads applied after composite action has been achieved) are determined as:





wD  10 ft  75 lb/ft 2  5 lb/ft 2  10 lb/ft 2  1 kip 1,000 lb     0.900 kip/ft





wL  10 ft  100 lb/ft 2  1 kip 1,000 lb     1.00 kip/ft

From ASCE/SEI 7, Chapter 2, the required flexural strength is: LRFD

ASD

wu  1.2  0.900 kip/ft   1.6 1.00 kip/ft   2.68 kip/ft w L2 Mu  u 8 

wa  0.900 kip/ft  1.00 kip/ft  1.90 kip/ft

Ma 

 2.68 kip/ft  45 ft 2



8

wa L2 8

1.90 kip/ft  45 ft 2 8

 481 kip-ft

 678 kip-ft

Determine effective width, b The effective width of the concrete slab is the sum of the effective widths to each side of the beam centerline as determined by the minimum value of the three widths set forth in AISC Specification Section I3.1a: 1.

one-eighth of the beam span, center-to-center of supports

45 ft  2 sides   11.3 ft 8 2.

one-half the distance to the centerline of the adjacent beam

10 ft  2 sides   10.0 ft controls 2

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3.

distance to the edge of the slab The latter is not applicable for an interior member.

Available Flexural Strength According to AISC Specification Section I3.2a, the nominal flexural strength shall be determined from the plastic stress distribution on the composite section when h / tw  3.76 E / Fy .

49.4  3.76

 29,000 ksi  /  50 ksi 

 90.6 Therefore, use the plastic stress distribution to determine the nominal flexural strength. According to the User Note in AISC Specification Section I3.2a, this check is generally unnecessary as all current W-shapes satisfy this limit for Fy  70 ksi.

Flexural strength can be determined using AISC Manual Table 3-19 or calculated directly using the provisions of AISC Specification Chapter I. This design example illustrates the use of the Manual table only. For an illustration of the direct calculation procedure, refer to Design Example I.2. To utilize AISC Manual Table 3-19, the distance from the compressive concrete flange force to beam top flange, Y2, must first be determined as illustrated by Manual Figure 3-3. Fifty percent composite action [Qn  0.50(AsFy)] is used to calculate a trial value of the compression block depth, atrial, for determining Y2 as follows: atrial   

 Qn 0.85 f cb

(from Manual Eq. 3-7)

0.50  As Fy  0.85 f cb





0.50 14.7 in.2  50 ksi  0.85  4 ksi 10 ft 12 in./ft 

 0.90 in.  say 1.00 in.

Note that a trial value of a = 1 in. is a common starting point in many design problems.

Y 2  Ycon 

atrial 2

(from Manual. Eq 3-6)

where

Ycon  distance from top of steel beam to top of slab, in.  7.50 in. Y 2  7.50 in. 

1 in. 2

 7.00 in.

Enter AISC Manual Table 3-19 with the required strength and Y2 = 7.00 in. to select a plastic neutral axis location for the W2150 that provides sufficient available strength.

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Selecting PNA location 5 (BFL) with  Qn  386 kips provides a flexural strength of: LRFD b M n  769 kip-ft  678 kip-ft

ASD

Mn  512 kip-ft  481 kip-ft o.k. b

o.k.

Based on the available flexural strength provided in Table 3-19, the required PNA location for ASD and LRFD design methodologies differ. This discrepancy is due to the live to dead load ratio in this example, which is not equal to the ratio of 3 at which ASD and LRFD design methodologies produce equivalent results as discussed in AISC Specification Commentary Section B3.2. The selected PNA location 5 is acceptable for ASD design, and more conservative for LRFD design. The actual value for the compression block depth, a, is determined as follows:

a 

 Qn 0.85 f cb

(Manual Eq. 3-7)

386 kips 0.85  4 ksi 10 ft 12 in./ft 

 0.946 in.  atrial  1.00 in. o.k. Live Load Deflection Deflections due to live load applied after composite action has been achieved will be limited to L / 360 under the design live load as required by Table 1604.3 of the International Building Code (IBC) (ICC, 2015), or 1 in. using a 50% reduction in design live load as recommended by AISC Design Guide 3. Deflections for composite members may be determined using the lower bound moment of inertia provided by Specification Commentary Equation C-I3-1 and tabulated in AISC Manual Table 3-20. The Specification Commentary also provides an alternate method for determining deflections of a composite member through the calculation of an effective moment of inertia. This design example illustrates the use of the Manual table. For an illustration of the direct calculation procedure for each method, refer to Design Example I.2.

Entering Table 3-20, for a W2150 with PNA location 5 and Y2 = 7.00 in., provides a lower bound moment of inertia of I LB  2, 520 in.4 Inserting ILB into AISC Manual Table 3-23, Case 1, to determine the live load deflection under the full design live load for comparison to the IBC limit yields: c  

5wL L4 384 EI LB 5 1.00 kip/ft 1 ft/12 in.  45 ft 12 in./ft  



384  29, 000 ksi  2,520 in.4

 1.26 in.  L / 429  L / 360

4



o.k.

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Performing the same check with 50% of the design live load for comparison to the AISC Design Guide 3 limit yields:  c  0.50 1.26 in.  0.630 in.  1 in. o.k. Steel Anchor Strength Steel headed stud anchor strengths are tabulated in AISC Manual Table 3-21 for typical conditions. Conservatively assuming that all anchors are placed in the weak position, the strength for w-in.-diameter anchors in normal weight concrete with f c  4 ksi and deck oriented perpendicular to the beam is: 1 anchor per rib: 2 anchors per rib:

Qn  17.2 kips/anchor Qn  14.6 kips/anchor

Number and Spacing of Anchors Deck flutes are spaced at 12 in. on center according to the deck manufacturer’s literature. The minimum number of deck flutes along each half of the 45-ft-long beam, assuming the first flute begins a maximum of 12 in. from the support line at each end, is:

n flutes  nspaces  1 

45 ft  2 12 in.1 ft/12 in. 2 1 ft per space 

1

 22.5  say 22 flutes According to AISC Specification Section I8.2c, the number of steel headed stud anchors required between the section of maximum bending moment and the nearest point of zero moment is determined by dividing the required horizontal shear,  Qn , by the nominal shear strength per anchor, Qn . Assuming one anchor per flute:  Qn Qn 386 kips  17.2 kips/anchor  22.4  place 23 anchors on each side of the beam centerline

nanchors 

As the number of anchors exceeds the number of available flutes by one, place two anchors in the first flute. The revised horizontal shear capacity of the anchors taking into account the reduced strength for two anchors in one flute is:  Qn  2 14.6 kips   2117.2 kips   390 kips  386 kips

o.k.

Steel Anchor Ductility Check As discussed in AISC Specification Commentary to Section I3.2d, beams are not susceptible to connector failure due to insufficient deformation capacity if they meet one or more of the following conditions: 1. 2.

Beams with span not exceeding 30 ft; Beams with a degree of composite action of at least 50%; or

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3.

Beams with an average nominal shear connector capacity of at least 16 kips per foot along their span, corresponding to a w-in.-diameter steel headed stud anchor placed at 12 in. spacing on average.

The span is 45 ft, which exceeds the 30 ft limit. The percent composite action is:  Qn 390 kips  min 0.85 f cAc , Fy As  min 0.85  4 ksi 10 ft 12 in./ft  4.5 in. ,  50 ksi  14.7 in.2







100 

390 kips 100  735 kips  53.1% 

which exceeds the minimum degree of composite action of 50%. The average shear connector capacity is:

 42 anchors 17.2 kips/anchor    4 anchors 14.6 kips/anchor  45 ft

 17.4 kip/ft

which exceeds the minimum capacity of 16 kips per foot. Since at least one of the conditions has been met (in fact, two have been met), the shear connectors meet the ductility requirements. The final anchor pattern chosen is illustrated in Figure I.1-2. Review steel headed stud anchor spacing requirements of AISC Specification Sections I8.2d and I3.2c. 1.

Maximum anchor spacing along beam [Section I8.2d(e)]: 8t slab  8  7.50 in.  60.0 in.

or 36 in. The maximum anchor spacing permitted is 36 in. 36 in.  12 in. o.k.

2.

Minimum anchor spacing along beam [Section I8.2d(d)]:

4d sa  4  w in.  3.00 in.  12 in. o.k. 3.

Minimum transverse spacing between anchor pairs [Section I8.2d(d)]: 4 d sa  4  w in.  3.00 in.  3.00 in.

4.

o.k.

Minimum distance to free edge in the direction of the horizontal shear force: AISC Specification Section I8.2d requires that the distance from the center of an anchor to a free edge in the direction of the shear force be a minimum of 8 in. for normal weight concrete slabs.

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Fig. I.1-2. Steel headed stud anchor layout. 5.

Maximum spacing of deck attachment: AISC Specification Section I3.2c.1(d) requires that steel deck be anchored to all supporting members at a maximum spacing of 18 in. The stud anchors are welded through the metal deck at a maximum spacing of 12 inches in this example, thus this limit is met without the need for additional puddle welds or mechanical fasteners.

Available Shear Strength According to AISC Specification Section I4.2, the beam should be assessed for available shear strength as a bare steel beam using the provisions of Chapter G. Applying the loads previously determined for the governing ASCE/SEI 7 load combinations and using available shear strengths from AISC Manual Table 3-2 for a W2150 yields the following: LRFD Vu  

wu L 2  2.68 kips/ft  45 ft 

Va  

2

 60.3 kips vVn  237 kips  60.3 kips

ASD wa L 2 1.90 kips/ft  45 ft  2

 42.8 kips

o.k.

Vn  158 kips  42.8 kips v

o.k.

Serviceability Depending on the intended use of this bay, vibrations might need to be considered. Refer to AISC Design Guide 11 (Murray et al., 2016) for additional information.

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Summary

From Figure I.1-2, the total number of stud anchors used is equal to (2)(2 + 21) = 46. A plan layout illustrating the final beam design is provided in Figure I.1-3. A W2150 with 2 in. of camber and 46, w-in.-diameter by 4d-in.long steel headed stud anchors is adequate to resist the imposed loads.

Fig. I.1-3. Revised plan.

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EXAMPLE I.2 COMPOSITE GIRDER DESIGN Given:

Two typical bays of a composite floor system are illustrated in Figure I.2-1. Select an appropriate ASTM A992 Wshaped girder and determine the required number of steel headed stud anchors. The girder will not be shored during construction. Use steel headed stud anchors made from ASTM A108 material, with Fu = 65 ksi.

Fig. I.2-1. Composite bay and girder section. To achieve a two-hour fire rating without the application of spray applied fire protection material to the composite deck, 42 in. of normal weight (145 lb/ft3) concrete will be placed above the top of the deck. The concrete has a specified compressive strength, f c = 4 ksi. Applied loads are given in the following: Dead Loads: Pre-composite: Slab = 75 lb/ft2 (in accordance with metal deck manufacturer’s data) Self-weight = 80 lb/ft (trial girder weight) = 50 lb/ft (beam weight from Design Example I.1) Composite (applied after composite action has been achieved): Miscellaneous = 10 lb/ft2 (HVAC, ceiling, floor covering, etc.) Live Loads: Pre-composite: Construction = 25 lb/ft2 (temporary loads during concrete placement)

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Composite (applied after composite action has been achieved): Non-reducible = 100 lb/ft2 (assembly occupancy) Solution:

From AISC Manual Table 2-4, the material properties are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi Applied Loads For slabs that are to be placed at a constant elevation, AISC Design Guide 3 (West and Fisher, 2003) recommends an additional 10% of the nominal slab weight be applied to account for concrete ponding due to deflections resulting from the wet weight of the concrete during placement. For the slab under consideration, this would result in an additional load of 8 lb/ft2; however, for this design the slab will be placed at a constant thickness, and thus, no additional weight for concrete ponding is required. For pre-composite construction live loading, 25 lb/ft2 will be applied in accordance with recommendations from Design Loads on Structures During Construction, ASCE/SEI 37 (ASCE, 2014), for a light duty operational class that includes concrete transport and placement by hose and finishing with hand tools. Composite Deck and Anchor Requirements Check composite deck and anchor requirements stipulated in AISC Specification Sections I1.3, I3.2c and I8. 3 ksi  f c  10 ksi (for normal weight concrete)

(Spec. Section I1.3)

1.

Concrete strength: f c  4 ksi o.k.

2.

Rib height: hr  3 in. hr  3 in. o.k.

(Spec. Section I3.2c)

3.

Average rib width: wr  2 in. wr  6 in. (See Figure I.2-1) o.k.

(Spec. Section I3.2c)

4.

Use steel headed stud anchors w in. or less in diameter.

(Spec. Section I8.1)

Select w-in.-diameter steel anchors. o.k. 5.

Steel headed stud anchor diameter: d sa  2.5t f

(Spec. Section I8.1)

In accordance with AISC Specification Section I8.1, this limit only applies if steel headed stud anchors are not welded to the flange directly over the web. The w-in.-diameter anchors will be attached in a staggered pattern, thus this limit must be satisfied. Select a girder size with a minimum flange thickness of 0.300 in., as determined in the following: d sa 2.5 w in.  2.5  0.300 in.

tf 

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6.

In accordance with AISC Specification I3.2c, steel headed stud anchors, after installation, shall extend not less than 12 in. above the top of the steel deck. A minimum anchor length of 42 in. is required to meet this requirement for 3-in.-deep deck. From steel headed stud anchor manufacturer’s data, a standard stock length of 4d in. is selected. Using a x-in. length reduction to account for burn off during anchor installation directly to the girder flange yields a final installed length of 4n in. 4n in. > 42 in. o.k.

7.

(Spec. Section I8.2)

Minimum length of stud anchors = 4dsa 4n in. > 4(w in.) = 3.00 in. o.k.

8.

In accordance with AISC Specification Section I3.2c, there shall be at least 2 in. of specified concrete cover above the top of the headed stud anchors. As discussed in the Specification Commentary to Section I3.2c, it is advisable to provide greater than 2-in. minimum cover to assure anchors are not exposed in the final condition. 72 in.  4n in.  2m in.  2 in. o.k.

9.

In accordance with AISC Specification Section I3.2c, slab thickness above steel deck shall not be less than 2 in. 42 in.  2 in.

o.k.

Design for Pre-Composite Condition Construction (Pre-Composite) Loads The girder will be loaded at third points by the supported beams. Determine point loads using tributary areas.









PD   45 ft 10 ft  75 lb/ft 2   45 ft  50 lb/ft   1 kip 1, 000 lb     36.0 kips PL   45 ft 10 ft  25 lb/ft 2  1 kip 1,000 lb     11.3 kips

Construction (Pre-Composite) Flexural Strength From ASCE/SEI 7, Chapter 2, the required flexural strength is: LRFD

Pu  1.2  36.0 kips   1.6 11.3 kips   61.3 kips

wu  1.2  80 lb/ft 1 kip 1, 000 lb   0.0960 kip/ft

ASD Pa  36.0 kips  11.3 kips  47.3 kips wa   80 lb/ft 1 kip 1, 000 lb   0.0800 kip/ft

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LRFD

M u  Pu a 

ASD

2

wu L 8

M a  Pa a 

  61.3 kips 10 ft  

 0.0960 kip/ft  30 ft 2 8

 624 kip-ft

2

wa L 8

  47.3 kips 10 ft  

 0.0800 kip/ft  30 ft 2 8

 482 kip-ft

Girder Selection Based on the required flexural strength under construction loading, a trial member can be selected utilizing AISC Manual Table 3-2. For the purposes of this example, the unbraced length of the girder prior to hardening of the concrete is taken as the distance between supported beams (one-third of the girder length). Try a W2476 Lb  10 ft L p  6.78 ft Lr  19.5 ft

LRFD

ASD

b M px  750 kip-ft

BF  b  15.1 kips M px  b  499 kip-ft

b M rx  462 kip-ft

M rx  b  307 kip-ft

b BF  22.6 kips

Because L p  Lb  Lr , use AISC Manual Equations 3-4a and 3-4b with Cb  1.0 within the center girder segment in accordance with AISC Manual Table 3-1:

LRFD

ASD

From AISC Manual Equation 3-4a:

From AISC Manual Equation 3-4b:

b M n  Cb  b M px  b BF ( Lb  L p )   b M px  1.0[750 kip-ft   22.6 kips  (10 ft  6.78 ft)]

Mn  M px BF  M px  Cb   ( Lb  L p )   b b b  b 

 750 kip-ft  677 kip-ft  750 kip-ft  677 kip-ft

b M n  M u 677 kip-ft  624 kip-ft

 1.0[499 kip-ft  15.1 kips 10 ft  6.78 ft ]

 499 kip-ft  450 kip-ft  499 kip-ft  450 kip-ft

o.k.

Mn  Ma b 450 kip-ft  482 kip-ft n.g.

For this example, the relatively low live load to dead load ratio results in a lighter member when LRFD methodology is employed. When ASD methodology is employed, a heavier member is required, and it can be shown that a W2484 is adequate for pre-composite flexural strength. This example uses a W2476 member to illustrate the determination of flexural strength of the composite section using both LRFD and ASD methodologies; however, this is done for comparison purposes only, and calculations for a W2484 would be required to provide a satisfactory ASD design. Calculations for the heavier section are not shown as they would essentially be a duplication of the calculations provided for the W2476 member.

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Note that for the member size chosen, 76 lb/ft < 80 lb/ft, thus the initial weight assumption is adequate.

From AISC Manual Table 1-1, the geometric properties are as follows: W2476 A = 22.4 in.2 h/tw = 49.0 Ix = 2,100 in.4 bf = 8.99 in. tf = 0.680 in. d = 23.9 in.

Pre-Composite Deflections AISC Design Guide 3 (West and Fisher, 2003) recommends deflections due to concrete plus self-weight not exceed the minimum of L/360 or 1.0 in. From the superposition of AISC Manual Table 3-23, Cases 1 and 9:  nc 

23PD L3 5wD L4  648 EI 384 EI

Substituting for the moment of inertia of the non-composite section, I  2,100 in.4 , yields a dead load deflection of:

 nc 

23  36.0 kips   30 ft 12 in./ft  



648  29, 000 ksi  2,100 in.4

3





5  0.0760 kip/ft 1 ft/12 in.  30 ft 12 in./ft  



384  29, 000 ksi  2,100 in.4

4



 1.00 in.  L / 360 o.k.

Pre-composite deflections barely meet the recommended value. Although technically acceptable, judgment leads one to consider ways to minimize pre-composite deflections. One possible solution is to increase the member size. A second solution is to introduce camber into the member. For this example, the second solution is selected, and the girder will be cambered to reduce pre-composite deflections. Reducing the estimated simple span deflections to 80% of the calculated value to reflect the partial restraint of the end connections as recommended in AISC Design Guide 3 yields a camber of: Camber = 0.80 1.00 in.  0.800 in.

Rounding down to the nearest 4-in. increment yields a specified camber of w in. Select a W2476 with w in. of camber.

Design for Composite Flexural Strength Required Flexural Strength Using tributary area calculations, the total applied point loads (including pre-composite dead loads in addition to dead and live loads applied after composite action has been achieved) are determined as:

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



PD   45 ft 10 ft  75 lb/ft 2  10 lb/ft 2   45 ft  50 lb/ft   1 kip 1, 000 lb     40.5 kips





PL   45 ft 10 ft  100 lb/ft 2  1 kip 1, 000 lb     45.0 kips

The required flexural strength diagram is illustrated by Figure I.2-2:

Fig. I.2-2. Required flexural strength. From ASCE/SEI 7, Chapter 2, the required flexural strength is: LRFD

ASD Pr  Pa  40.5 kips  45.0 kips  85.5 kips

Pr  Pu  1.2  40.5 kips   1.6  45.0 kips   121 kips

wu  1.2  0.0760 kip/ft 

wa  0.0760 kip/ft (from self weight of W24×76)

 0.0912 kip/ft (from self weight of W24×76) LRFD From AISC Manual Table 3-23, Case 1 and 9:

ASD From AISC Manual Table 3-23, Case 1 and 9:

M u1  M u 3

M a1  M a 3

wu a  L  a 2  121 kips 10 ft 

wa a  L  a 2   85.5 kips 10 ft 

 Pu a 



 0.0912 kip/ft 10 ft 

 1, 220 kip-ft

2

 Pa a 

 30 ft  10 ft 



 0.0760 kip/ft 10 ft 

 863 kip-ft

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 30 ft  10 ft 

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LRFD M u2

ASD

w L2  Pu a  u 8  121 kips 10 ft  

M a2

 0.0912 kip/ft  30 ft 2 8

 1, 220 kip-ft

w L2  Pa a  a 8   85.5 kips 10 ft  

 0.0760 kip/ft  30 ft 2 8

 864 kip-ft

Determine Effective Width, b The effective width of the concrete slab is the sum of the effective widths to each side of the beam centerline as determined by the minimum value of the three conditions set forth in AISC Specification Section I3.1a: 1.

one-eighth of the girder span center-to-center of supports

 30 ft     2 sides   7.50 ft controls  8  2.

one-half the distance to the centerline of the adjacent girder  45 ft     2 sides   45.0 ft  2 

3.

distance to the edge of the slab The latter is not applicable for an interior member.

Available Flexural Strength

According to AISC Specification Section I3.2a, the nominal flexural strength shall be determined from the plastic stress distribution on the composite section when h / tw  3.76 E / Fy . 49.0  3.76

29, 000 ksi 50 ksi

 90.6

Therefore, use the plastic stress distribution to determine the nominal flexural strength. According to the User Note in AISC Specification Section I3.2a, this check is generally unnecessary as all current W-shapes satisfy this limit for Fy  70 ksi.

AISC Manual Table 3-19 can be used to facilitate the calculation of flexural strength for composite beams. Alternately, the available flexural strength can be determined directly using the provisions of AISC Specification Chapter I. Both methods will be illustrated for comparison in the following calculations. Method 1: AISC Manual

To utilize AISC Manual Table 3-19, the distance from the compressive concrete flange force to beam top flange, Y2, must first be determined as illustrated by Manual Figure 3-3. Fifty percent composite action [Qn  0.50(AsFy)] is used to calculate a trial value of the compression block depth, atrial, for determining Y2 as follows:

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atrial   

 Qn 0.85 f cb

(from Manual Eq. 3-7)

0.50  As Fy  0.85 f cb





0.50 22.4 in.2  50 ksi  0.85  4 ksi  7.50 ft 12 in./ft 

 1.83 in.

Y 2  Ycon 

atrial 2

(from Manual. Eq. 3-6)

where

Ycon  distance from top of steel beam to top of slab  7.50 in. Y 2  7.50 in. 

1.83 in. 2

 6.59 in. Enter AISC Manual Table 3-19 with the required strength and Y 2  6.59 in. to select a plastic neutral axis location for the W2476 that provides sufficient available strength. Based on the available flexural strength provided in Table 3-19, the required PNA location for ASD and LRFD design methodologies differ. This discrepancy is due to the live-to-dead load ratio in this example, which is not equal to the ratio of 3 at which ASD and LRFD design methodologies produce equivalent results as discussed in AISC Specification Commentary Section B3.2. Selecting PNA location 5 (BFL) with  Qn  509 kips provides a flexural strength of: LRFD b M n  1, 240 kip-ft  1, 220 kip-ft

ASD o.k.

Mn  823 kip-ft  864 kip-ft b

n.g.

The selected PNA location 5 is acceptable for LRFD design, but inadequate for ASD design. For ASD design, it can be shown that a W2476 is adequate if a higher composite percentage of approximately 60% is employed. However, as discussed previously, this beam size is not adequate for construction loading and a larger section is necessary when designing utilizing ASD. The actual value for the compression block depth, a, for the chosen PNA location is determined as follows:

a 

 Qn 0.85 f cb

(Manual Eq. 3-7)

509 kips 0.85  4 ksi  7.50 ft 12 in./ft 

 1.66 in.  atrial  1.83 in. o.k. for LRFD design Method 2: Direct Calculation

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According to AISC Specification Commentary Section I3.2a, the number and strength of steel headed stud anchors will govern the compressive force, C, for a partially composite beam. The composite percentage is based on the minimum of the limit states of concrete crushing and steel yielding as follows: 1.

Concrete crushing

Ac  Area of concrete slab within effective width. Assume that the deck profile is 50% void and 50% concrete fill.  beff  42 in.   beff / 2   3 in.   7.50 ft 12 in./ft     7.50 ft 12 in./ft  42 in.     3 in. 2    540 in.2 C  0.85 f cAc



 0.85  4 ksi  540 in.

2

(Spec. Comm. Eq. C-I3-7)



 1,840 kips

2.

Steel yielding C  As Fy



(Spec. Comm. Eq. C-I3-6) 2

 22.4 in.

 50 ksi 

 1,120 kips

3.

Shear transfer Fifty percent is used as a trial percentage of composite action as follows: C  Qn

(Spec. Comm. Eq. C-I3-8)

 1,840 kips    50%  min   1,120 kips     560 kips to achieve 50% composite action

Location of the Plastic Neutral Axis The plastic neutral axis (PNA) is located by determining the axis above and below which the sum of horizontal forces is equal. This concept is illustrated in Figure I.2-3, assuming the trial PNA location is within the top flange of the girder.

F above PNA  F below PNA

C  xb f Fy   As  b f x  Fy Solving for x:

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x

As Fy  C 2b f Fy

 22.4 in.  50 ksi   560 kips  2

2  8.99 in. 50 ksi 

 0.623 in.  t f  0.680 in.; therefore, the PNA is in the flange

Determine the nominal moment resistance of the composite section following the procedure in AISC Specification Commentary Section I3.2a, as illustrated in Figure C-I3.3.

a 

C 0.85 f cb

(Spec. Comm. Eq. C-I3-9)

560 kips 0.85  4 ksi  7.50 ft 12 in./ft 

 1.83 in.< 4.50 in. (above top of deck) d1  tslab 

a 2

 7.50 in. 

1.83 in. 2

 6.59 in. x 2 0.623 in.  2  0.312 in.

d2 

Fig. I.2-3. Plastic neutral axis location.

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d 2 23.9 in.  2  12.0 in.

d3 

Py  As Fy





 22.4 in.2  50 ksi   1,120 kips

M n  C  d1  d 2   Py  d3  d 2 

(Spec. Comm. Eq. C-I3-10)

  560 kips  6.59 in.  0.312 in.  1,120 kips 12.0 in.  0.312 in.  17, 000 kip-in. or 1,420 kip-ft Note that Equation C-I3-10 is based on the summation of moments about the centroid of the compression force in the steel; however, the same answer may be obtained by summing moments about any arbitrary point. LRFD

ASD

b  0.90

 b  1.67

b M n  0.90 1, 420 kip-ft 

M n 1, 420 kip-ft  1.67 b  850 kip-ft  864 kip-ft n.g.

 1, 280 kip-ft  1, 220 kip-ft

o.k.

As was determined previously using the Manual Tables, a W2476 with 50% composite action is acceptable when LRFD methodology is employed, while for ASD design the beam is inadequate at this level of composite action. Continue with the design using a W2476 with 50% composite action.

Steel Anchor Strength Steel headed stud anchor strengths are tabulated in AISC Manual Table 3-21 for typical conditions and may be calculated according to AISC Specification Section I8.2a as follows: Asa  

2 d sa 4

  w in.

2

4  0.442 in.2 f c  4 ksi

Ec  wc1.5 f c



 145 lb/ft 3



1.5

4 ksi

 3, 490 ksi

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Rg  1.0, stud anchors welded directly to the steel shape within the slab haunch Rp  0.75, stud anchors welded directly to the steel shape Fu  65 ksi

Qn  0.5 Asa



f cEc  Rg R p Asa Fu

  0.5  0.442 in.

2



(Spec. Eq. I8-1)

 4 ksi  3, 490 ksi   1.0  0.75   0.442 in.   65 ksi  2

 26.1 kips  21.5 kips

Use Qn = 21.5 kips. Number and Spacing of Anchors According to AISC Specification Section I8.2c, the number of steel headed stud anchors required between any concentrated load and the nearest point of zero moment shall be sufficient to develop the maximum moment required at the concentrated load point. From Figure I.2-2 the moment at the concentrated load points, Mr1 and Mr3, is approximately equal to the maximum beam moment, Mr2. The number of anchors between the beam ends and the point loads should therefore be adequate to develop the required compressive force associated with the maximum moment, C, previously determined to be 560 kips.  Qn Qn C  Qn 560 kips  21.5 kips/anchor

N anchors 

 26 anchors from each end to concentrated load points

In accordance with AISC Specification Section I8.2d, anchors between point loads should be spaced at a maximum of: 8tslab  60.0 in. or 36 in. controls For beams with deck running parallel to the span such as the one under consideration, spacing of the stud anchors is independent of the flute spacing of the deck. Single anchors can therefore be spaced as needed along the beam length provided a minimum longitudinal spacing of six anchor diameters in accordance with AISC Specification Section I8.2d is maintained. Anchors can also be placed in aligned or staggered pairs provided a minimum transverse spacing of four stud diameters = 3 in. is maintained. For this design, it was chosen to use pairs of anchors along each end of the girder to meet strength requirements and single anchors along the center section of the girder to meet maximum spacing requirements as illustrated in Figure I.2-4. AISC Specification Section I8.2d requires that the distance from the center of an anchor to a free edge in the direction of the shear force be a minimum of 8 in. for normal weight concrete slabs. For simply-supported composite beams this provision could apply to the distance between the slab edge and the first anchor at each end of the beam. Assuming the slab edge is coincident to the centerline of support, Figure I.2-4 illustrates an acceptable edge distance of 9 in., though in this case the column flange would prevent breakout and negate the need for this check. The slab edge is often uniformly supported by a column flange or pour stop in typical composite construction thus preventing

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the possibility of a concrete breakout failure and nullifying the edge distance requirement as discussed in AISC Specification Commentary Section I8.3. For this example, the minimum number of headed stud anchors required to meet the maximum spacing limit previously calculated is used within the middle third of the girder span. Note also that AISC Specification Section I3.2c.1(d) requires that steel deck be anchored to all supporting members at a maximum spacing of 18 in. Additionally, Standard for Composite Steel Floor Deck-Slabs, ANSI/SDI C1.0-2011 (SDI, 2011), requires deck attachment at an average of 12 in. but no more than 18 in. From the previous discussion and Figure I.2-4, the total number of stud anchors used is equal to 13 2   3  13 2   55 . A plan layout illustrating the final girder design is provided in Figure I.2-5. Steel Anchor Ductility Check As discussed in AISC Specification Commentary Section I3.2d, beams are not susceptible to connector failure due to insufficient deformation capacity if they meet one or more of the following conditions: (1) Beams with span not exceeding 30 ft; (2) Beams with a degree of composite action of at least 50%; or

Fig. I.2-4. Steel headed stud anchor layout.

Fig. I.2-5. Revised plan.

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(3) Beams with an average nominal shear connector capacity of at least 16 kips per foot along their span, corresponding to a w-in.-diameter steel headed stud anchor placed at 12-in. spacing on average. The span is 30 ft, which meets the 30 ft limit. The percent composite action is:  Qn 560 kips  min 0.85 f cAc , Fy As  min 0.85  4 ksi  540 in.2 ,  50 ksi  22.4 in.2











560 kips 100  1,120 kips  50.0% 

which meets the minimum degree of composite action of 50%. The average shear connector capacity is:

 55 anchors  21.5 kips/anchor  30 ft

 39.4 kip/ft

which exceeds the minimum capacity of 16 kips per foot. Because at least one of the conditions has been met (in fact, all three have been met), the shear connectors meet the ductility requirements. Live Load Deflection Criteria Deflections due to live load applied after composite action has been achieved will be limited to L / 360 under the design live load as required by Table 1604.3 of the International Building Code (IBC) (ICC, 2015), or 1 in. using a 50% reduction in design live load as recommended by AISC Design Guide 3. Deflections for composite members may be determined using the lower bound moment of inertia provided in AISC Specification Commentary Equation C-I3-1 and tabulated in AISC Manual Table 3-20. The Specification Commentary also provides an alternate method for determining deflections through the calculation of an effective moment of inertia. Both methods are acceptable and are illustrated in the following calculations for comparison purposes:

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Method 1: Calculation of the lower bound moment of inertia, ILB 2   Qn I LB  I x  As YENA  d3     Fy

 2   2d3  d1  YENA  

(Spec. Comm. Eq. C-I3-1)

Variables d1 and d3 in AISC Specification Commentary Equation C-I3-1 are determined using the same procedure previously illustrated for calculating nominal moment resistance. However, for the determination of I LB the nominal strength of steel anchors is calculated between the point of maximum positive moment and the point of zero moment as opposed to between the concentrated load and point of zero moment used previously. The maximum moment is located at the center of the span and it can be seen from Figure I.2-4 that 27 anchors are located between the midpoint of the beam and each end. Qn   27 anchors  21.5 kips/anchor   581 kips C 0.85 f cb  Qn  0.85 f cb

a



(Spec. Eq. C-I3-9)

581 kips 0.85  4 ksi  7.50 ft 12 in./ft 

 1.90 in. d1  tslab 

a 2

 7.50 in. 

1.90 in. 2

 6.55 in.

x=

As Fy   Qn 2b f Fy

 22.4 in.  50 ksi   581 kips  2

2  8.99 in. 50 ksi 

 0.600 in.  t f  0.680 in.; therefore, the PNA is within the flange d 2 23.9 in.  2  12.0 in.

d3 

The distance from the top of the steel section to the elastic neutral axis, YENA, for use in Equation C-I3-1 is calculated using the procedure provided in AISC Specification Commentary Section I3.2 as follows:

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YENA

  Qn  As d3     2d3  d1  Fy      Qn  As     Fy 

(Spec. Comm. Eq. C-I3-2)

kips   22.4 in.  12.0 in.   581   2 12.0 in.  6.55 in. 50 ksi   2



 581 kips  22.4 in.2     50 ksi 

 18.3 in.

Substituting these values into AISC Specification Commentary Equation C-I3-1 yields the following lower bound moment of inertia: 2 2  581 kips  I LB  2,100 in.4  22.4 in.2 18.3 in.  12.0 in.     2 12.0 in.  6.55 in.  18.3 in.  50 ksi 





 4, 730 in.4 Alternately, this value can be determined directly from AISC Manual Table 3-20 as illustrated in Design Example I.1. Method 2: Calculation of the equivalent moment of inertia, Iequiv An alternate procedure for determining a moment of inertia for the deflection calculation of the composite section is presented in AISC Specification Commentary Section I3.2 and in the following:

Determine the transformed moment of inertia, Itr The effective width of the concrete below the top of the deck may be approximated with the deck profile resulting in a 50% effective width as depicted in Figure I.2-6. The effective width, beff = (7.50 ft)(12 in./ft) = 90.0 in.

Transformed slab widths are calculated as follows: Es Ec 29, 000 ksi  3, 490 ksi

n

 8.31 btr1 

beff

n 90.0 in.  8.31  10.8 in.

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btr 2  

0.5beff n 0.5  90.0 in.

8.31  5.42 in.

The transformed model is illustrated in Figure I.2-7.

Determine the elastic neutral axis of the transformed section (assuming fully composite action) and calculate the transformed moment of inertia using the information provided in Table I.2-1 and Figure I.2-7. For this problem, a trial location for the elastic neutral axis (ENA) is assumed to be within the depth of the composite deck. Table I.2-1. Properties for Elastic Neutral Axis Determination of Transformed Section y, I, A, Part 2 4 in. in. in. 2.25  x A1 48.6 82.0 A2 5.42x x/2 0.452x3 W2476 22.4 x  15.0 2,100

Fig. I.2-6. Effective concrete width.

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Fig. I.2-7. Transformed area model. Ay about elastic neutral axis  0 

2



 48.6 in.   2.25 in.  x   5.42 in.  x2    22.4 in.   x  15.0 in.  0 2



2



Solving for x: x  2.88 in.

Verify trial location: 2.88 in.  hr  3 in.; therefore, the elastic neutral axis is within the composite deck

Utilizing the parallel axis theorem and substituting for x yields: I tr  I  Ay 2



 82.0 in.4   0.452 in. 2.88 in.  2,100 in.4  48.6 in.2 3



 22.4 in.2

  2.88 in.  15.0 in.

  2.25 in.  2.88 in.  15.6 in.   2.882 in.  2

2

2

2

 6,800 in.4 Determine the equivalent moment of inertia, Iequiv Qn  581 kips (previously determined in Method 1)

C f  compression force for fully composite beam previously determined to be controlled by As Fy  1,120 kips

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I equiv  I s 

 Qn / C f   Itr  I s 

 2,100 in.4 

(Spec. Comm. Eq. C-I3-3)

 581 kips) / (1,120 kips   6,800 in.4  2,100 in.4 

 5, 490 in.4 Comparison of Methods and Final Deflection Calculation

ILB was determined to be 4,730 in.4 and Iequiv was determined to be 5,490 in.4 ILB will be used for the remainder of this example. From AISC Manual Table 3-23, Case 9:

 LL 

23PL L3 648 EI LB

23  45.0 kips   30 ft 12 in./ft    648  29, 000 ksi  4, 730 in.4



3



 0.543 in.  1.00 in. (for AISC Design Guide 3 limit)

o.k.

(50% reduction in design live load as allowed by Design Guide 3 was not necessary to meet this limit)  L / 662  L / 360 (for IBC 2015 Table 1604.3 limit) o.k. Available Shear Strength According to AISC Specification Section I4.2, the girder should be assessed for available shear strength as a bare steel beam using the provisions of Chapter G. Applying the loads previously determined for the governing load combination of ASCE/SEI 7 and obtaining available shear strengths from AISC Manual Table 3-2 for a W2476 yields the following: LRFD

ASD

 30 ft  Vu  121 kips   0.0912 kip/ft     2   122 kips

 30 ft  Va  85.5 kips   0.0760 kip/ft     2   86.6 kips

vVn  315 kips  122 kips

Vn  210 kips  86.6 kips o.k. v

o.k.

Serviceability Depending on the intended use of this bay, vibrations might need to be considered. See AISC Design Guide 11 (Murray et al., 2016) for additional information. It has been observed that cracking of composite slabs can occur over girder lines. The addition of top reinforcing steel transverse to the girder span will aid in mitigating this effect. Summary Using LRFD design methodology, it has been determined that a W2476 with w in. of camber and 55, w-in.diameter by 4d-in.-long steel headed stud anchors as depicted in Figure I.2-4, is adequate for the imposed loads and deflection criteria. Using ASD design methodology, a W2484 with a steel headed stud anchor layout determined using a procedure analogous to the one demonstrated in this example would be required.

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EXAMPLE I.3 FILLED COMPOSITE MEMBER FORCE ALLOCATION AND LOAD TRANSFER Given: Refer to Figure I.3-1. Part I: For each loading condition (a) through (c) determine the required longitudinal shear force, Vr , to be transferred between the steel section and concrete fill. Part II: For loading condition (a), investigate the force transfer mechanisms of direct bearing, shear connection, and direct bond interaction. The composite member consists of an ASTM A500, Grade C, HSS with normal weight (145 lb/ft3) concrete fill having a specified concrete compressive strength, f c = 5 ksi. Use ASTM A36 material for the bearing plate. Applied loading, Pr, for each condition illustrated in Figure I.3-1 is composed of the following nominal loads: PD = 32 kips PL = 84 kips

Fig. I.3-1. Filled composite member in compression.

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Solution: Part I—Force Allocation From AISC Manual Table 2-4, the material properties are as follows: ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi From AISC Manual Table 1-11 and Figure I.3-1, the geometric properties are as follows: HSS106a

As H B tnom t h/t b/t

= 10.4 in.2 = 10.0 in. = 6.00 in. = a in. (nominal wall thickness) = 0.349 in. (design wall thickness in accordance with AISC Specification Section B4.2) = 25.7 = 14.2

Calculate the concrete area using geometry compatible with that used in the calculation of the steel area in AISC Manual Table 1-11 (taking into account the design wall thickness and an outside corner radii of two times the design wall thickness in accordance with AISC Manual Part 1), as follows: hi  H  2t  10.0 in.  2  0.349 in.  9.30 in. bi  B  2t  6.00 in.  2  0.349 in.  5.30 in.

Ac  bi hi  t 2  4      5.30 in. 9.30 in.   0.349 

2

 4  

2

 49.2 in.

From ASCE/SEI 7, Chapter 2, the required compressive strength is: LRFD

ASD Pr  Pa  32 kips  84 kips  116 kips

Pr  Pu  1.2  32 kips   1.6  84 kips   173 kips Composite Section Strength for Force Allocation

In order to determine the composite section strength for force allocation, the member is first classified as compact, noncompact or slender in accordance with AISC Specification Table I1.1a.

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Governing Width-to-Thickness Ratio 

h t  25.7



The limiting width-to-thickness ratio for a compact compression steel element in a composite member subject to axial compression is:   p  2.26

E Fy

(Spec. Table I1.1a)

29,000 ksi 50 ksi  54.4  25.7; therefore the HSS wall is compact  2.26

The nominal axial compressive strength without consideration of length effects, Pno, used for force allocation calculations is therefore determined as:

Pno  Pp

(Spec. Eq. I2-9a)

E   Pp  Fy As  C2 f c  Ac  Asr s  Ec  

(Spec. Eq. I2-9b)

where C2 = 0.85 for rectangular sections Asr = 0 in.2 when no reinforcing steel is present within the HSS E   Pno  Fy As  C2 f c  Ac  Asr s  Ec  







  50 ksi  10.4 in.2  0.85  5 ksi  49.2 in.2  0 in.2



 729 kips Transfer Force for Condition (a) Refer to Figure I.3-1(a). For this condition, the entire external force is applied to the steel section only, and the provisions of AISC Specification Section I6.2a apply.

 Fy As  Vr  Pr 1   Pno  

(Spec. Eq. I6-1)



  50 ksi  10.4 in.2  Pr 1   729 kips   0.287 Pr

   

LRFD

Vr  0.287 173 kips   49.7 kips

ASD

Vr  0.287 116 kips   33.3 kips

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Transfer Force for Condition (b) Refer to Figure I.3-1(b). For this condition, the entire external force is applied to the concrete fill only, and the provisions of AISC Specification Section I6.2b apply.  Fy As  Vr  Pr    Pno    50 ksi  10.4 in.2  Pr   729 kips   0.713Pr



(Spec. Eq. I6-2a)

   

LRFD

Vr  0.713 173 kips 

ASD

Vr  0.713 116 kips 

 123 kips

 82.7 kips

Transfer Force for Condition (c) Refer to Figure I.3-1(c). For this condition, external force is applied to the steel section and concrete fill concurrently, and the provisions of AISC Specification Section I6.2c apply. AISC Specification Commentary Section I6.2 states that when loads are applied to both the steel section and concrete fill concurrently, Vr can be taken as the difference in magnitudes between the portion of the external force applied directly to the steel section and that required by Equation I6-2a and b. Using the plastic distribution approach employed in AISC Specification Equations I6-1 and I6-2a, this concept can be written in equation form as follows:

 As Fy  Vr  Prs  Pr    Pno 

(Eq. 1)

where Prs = portion of external force applied directly to the steel section, kips Note that this example assumes the external force imparts compression on the composite element as illustrated in Figure I.3-1. If the external force would impart tension on the composite element, consult the AISC Specification Commentary for discussion. Currently the Specification provides no specific requirements for determining the distribution of the applied force for the determination of Prs, so it is left to engineering judgment. For a bearing plate condition such as the one represented in Figure I.3-1(c), one possible method for determining the distribution of applied forces is to use an elastic distribution based on the material axial stiffness ratios as follows: Ec  wc1.5 f c



 145 lb/ft 3



1.5

5 ksi

 3,900 ksi

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Es As  Prs    Es As  Ec Ac

  Pr 





  29, 000 ksi  10.4 in.2     29, 000 ksi  10.4 in.2   3,900 ksi  49.2 in.2   0.611Pr









  Pr  

Substituting the results into Equation 1 yields:  As Fy  Vr  0.611Pr  Pr    Pno 





 10.4 in.2  50 ksi     0.611Pr  Pr    729 kips    0.102 Pr LRFD

Vr  0.102 173 kips   17.6 kips

ASD

Vr  0.102 116 kips   11.8 kips

An alternate approach would be the use of a plastic distribution method whereby the load is partitioned to each material in accordance with their contribution to the composite section strength given in Equation I2-9b. This method eliminates the need for longitudinal shear transfer provided the local bearing strength of the concrete and steel are adequate to resist the forces resulting from this distribution. Additional Discussion



The design and detailing of the connections required to deliver external forces to the composite member should be performed according to the applicable sections of AISC Specification Chapters J and K. Note that for checking bearing strength on concrete confined by a steel HSS or box member, the A2 / A1 term in Equation J8-2 may be taken as 2.0 according to the User Note in Specification Section I6.2.



The connection cases illustrated by Figure I.3-1 are idealized conditions representative of the mechanics of actual connections. For instance, a standard shear connection welded to the face of an HSS column is an example of a condition where all external force is applied directly to the steel section only. Note that the connection configuration can also impact the strength of the force transfer mechanism as illustrated in Part II of this example.

Solution: Part II—Load Transfer The required longitudinal force to be transferred, Vr , determined in Part I condition (a) will be used to investigate the three applicable force transfer mechanisms of AISC Specification Section I6.3: direct bearing, shear connection, and direct bond interaction. As indicated in the Specification, these force transfer mechanisms may not be superimposed; however, the mechanism providing the greatest nominal strength may be used.

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Direct Bearing Trial Layout of Bearing Plate For investigating the direct bearing load transfer mechanism, the external force is delivered directly to the HSS section by standard shear connections on each side of the member as illustrated in Figure I.3-2. One method for utilizing direct bearing in this instance is through the use of an internal bearing plate. Given the small clearance within the HSS section under consideration, internal access for welding is limited to the open ends of the HSS; therefore, the HSS section will be spliced at the bearing plate location. Additionally, it is a practical consideration that no more than 50% of the internal width of the HSS section be obstructed by the bearing plate in order to facilitate concrete placement. It is essential that concrete mix proportions and installation of concrete fill produce full bearing above and below the projecting plate. Based on these considerations, the trial bearing plate layout depicted in Figure I.3-2 was selected using an internal plate protrusion, Lp, of 1.0 in. Location of Bearing Plate The bearing plate is placed within the load introduction length discussed in AISC Specification Section I6.4b. The load introduction length is defined as two times the minimum transverse dimension of the HSS both above and below the load transfer region. The load transfer region is defined in Specification Commentary Section I6.4 as the depth of the connection. For the configuration under consideration, the bearing plate should be located within 2(B = 6 in.) = 12 in. of the bottom of the shear connection. From Figure I.3-2, the location of the bearing plate is 6 in. from the bottom of the shear connection and is therefore adequate. Available Strength for the Limit State of Direct Bearing The contact area between the bearing plate and concrete, A1, may be determined as follows:

A1  Ac  (bi  2 L p )(hi  2 L p )

(Eq. 2)

where L p  typical protrusion of bearing plate inside HSS

 1.0 in.

Fig. I.3-2. Internal bearing plate configuration.

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Substituting for the appropriate geometric properties previously determined in Part I into Equation 2 yields: A1  49.2 in.2  5.30 in.  2 1.0 in.  9.30 in.  2 1.0 in.   25.1 in.2

The available strength for the direct bearing force transfer mechanism is: Rn  1.7 f cA1

(Spec. Eq. I6-3) LRFD

ASD

B  0.65

B  2.31



B Rn  0.65 1.7  5 ksi  25.1 in.2  139 kips  Vr  49.7 kips







2 Rn 1.7  5 ksi  25.1 in.  2.31 B  92.4 kips  Vr  33.3 kips o.k.

o.k.

Required Thickness of Internal Bearing Plate There are several methods available for determining the bearing plate thickness. For round HSS sections with circular bearing plate openings, a closed-form elastic solution such as those found in Roark’s Formulas for Stress and Strain (Young and Budynas, 2002) may be used. Alternately, the use of computational methods such as finite element analysis may be employed. For this example, yield line theory can be employed to determine a plastic collapse mechanism of the plate. In this case, the walls of the HSS lack sufficient stiffness and strength to develop plastic hinges at the perimeter of the bearing plate. Utilizing only the plate material located within the HSS walls, and ignoring the HSS corner radii, the yield line pattern is as depicted in Figure I.3-3.

Fig. I.3-3. Yield line pattern.

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Utilizing the results of the yield line analysis with Fy  36 ksi plate material, the plate thickness may be determined as follows: ASD

LRFD   0.90

tp 

  1.67

8L p 2  wu   L p  bi  hi    Fy  3 

tp 

where wu  bearing pressure on plate determined using LRFD load combinations V  r A1 49.7 kips  25.1 in.2  1.98 ksi

tp 

  1.98 ksi      0.90  36 ksi  

2  8 1.0 in.    1.0 in. 5.30 in.  9.30 in.  3    0.604 in.

 wa  Fy

 8L p 2   L p  bi  hi    3  

where wa  bearing pressure on plate determined using ASD load combinations V  r A1 33.3 kips  25.1 in.2  1.33 ksi

tp 

 1.67 1.33 ksi        36 ksi  

2  8 1.0 in.    1.0 in. 5.30 in.  9.30 in.  3    0.607 in.

Thus, select a w-in.-thick bearing plate. Splice Weld The HSS is in compression due to the imposed loads, therefore the splice weld indicated in Figure I.3-2 is sized according to the minimum weld size requirements of Chapter J. Should uplift or flexure be applied in other loading conditions, the splice should be designed to resist these forces using the applicable provisions of AISC Specification Chapters J and K. Shear Connection Shear connection involves the use of steel headed stud or channel anchors placed within the HSS section to transfer the required longitudinal shear force. The use of the shear connection mechanism for force transfer in filled HSS is usually limited to large HSS sections and built-up box shapes, and is not practical for the composite member in question. Consultation with the fabricator regarding their specific capabilities is recommended to determine the feasibility of shear connection for HSS and box members. Should shear connection be a feasible load transfer mechanism, AISC Specification Section I6.3b in conjunction with the steel anchors in composite component provisions of Section I8.3 apply. Direct Bond Interaction The use of direct bond interaction for load transfer is limited to filled HSS and depends upon the location of the load transfer point within the length of the member being considered (end or interior) as well as the number of faces to which load is being transferred.

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From AISC Specification Section I6.3c, the nominal bond strength for a rectangular section is: Rn  pb Lin Fin

(Spec. Eq. I6-5)

where pb = perimeter of the steel-concrete bond interface within the composite cross section, in.    0.349 in.  =  2 10.0 in.  6.00 in.   8   2  0.349 in.    4    2    28.6 in.

Lin  load introduction length, determined in accordance with AISC Specification Section I6.4  2  min  B, H    2  6.00 in.  12.0 in. Fin 

12t

 0.1, ksi (for a rectangular cross section) H2 12  0.349 in. =  0.1 ksi 10.0 in.2  0.0419 ksi

For the design of this load transfer mechanism, two possible cases will be considered: Case 1: End Condition—Load Transferred to Member from Four Sides Simultaneously For this case the member is loaded at an end condition (the composite member only extends to one side of the point of force transfer). Force is applied to all four sides of the section simultaneously thus allowing the full perimeter of the section to be mobilized for bond strength. From AISC Specification Equation I6-5: LRFD

ASD

  0.50

  3.00

Rn  pb Lin Fin

Rn pb Lin Fin     28.6 in.12.0 in. 0.0419 ksi   3.00   4.79 kips  Vr  33.3 kips n.g.

 0.50  28.6 in.12.0 in. 0.0419 ksi   7.19 kips  Vr  49.7 kips

n.g.

Bond strength is inadequate and another force transfer mechanism such as direct bearing must be used to meet the load transfer provisions of AISC Specification Section I6. Alternately, the detail could be revised so that the external force is applied to both the steel section and concrete fill concurrently as schematically illustrated in Figure I.3-1(c). Comparing bond strength to the load transfer requirements for concurrent loading determined in Part I of this example yields:

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LRFD

ASD

  3.00

  0.50 Rn  7.19 kips  Vr  17.6 kips

n.g.

Rn  4.79 kips  Vr  11.8 kips n.g. 

Bond strength remains inadequate and another force transfer mechanism such as direct bearing must be used to meet the load transfer provisions of AISC Specification Section I6. Case 2: Interior Condition—Load Transferred to Three Faces For this case the composite member is loaded from three sides away from the end of the member (the composite member extends to both sides of the point of load transfer) as indicated in Figure I.3-4.

Fig. I.3-4. Case 2 load transfer. Longitudinal shear forces to be transferred at each face of the HSS are calculated using the relationship to external forces determined in Part I of the example for condition (a) shown in Figure I.3-1, and the applicable ASCE/SEI 7 load combinations as follows: LRFD Face 1: Pr1  Pu

 1.2  2 kips   1.6  6 kips   12.0 kips Vr1  0.287 Pr1  0.287 12.0 kips   3.44 kips

ASD Face 1: Pr1  Pa  2 kips  6 kips  8.00 kips Vr1  0.287 Pr1  0.287  8.00 kips   2.30 kips

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LRFD

ASD Faces 2 and 3: Pr 2 3  Pu  15 kips  39 kips  54.0 kips

Faces 2 and 3: Pr 23  Pu

 1.2 15 kips   1.6  39 kips   80.4 kips

Vr2 3  0.287 Pr 2 3

Vr2 3  0.287 Pr 23

 0.287  54.0 kips 

 0.287  80.4 kips 

 15.5 kips

 23.1 kips

Load transfer at each face of the section is checked separately for the longitudinal shear at that face using Equation I6-5 as follows: LRFD

ASD

  0.50

  3.00

Face 1: pb  6.00 in.   2 corners  2  0.349 in.

Face 1: pb  6.00 in.   2 corners  2  0.349 in.  4.60 in.

 4.60 in.

 1.16 kips  Vr1  3.44 kips n.g.

Rn1  4.60 in.12.0 in. 0.0419 ksi   3.00   0.771 kip  Vr1  2.30 kips n.g.

Faces 2 and 3: pb  10.0 in.   2 corners  2  0.349 in.

Faces 2 and 3: pb  10.0 in.   2 corners  2  0.349 in.

Rn1  0.50  4.60 in.12.0 in. 0.0419 ksi 

 8.60 in.

 8.60 in.

Rn 23  0.50  8.60 in.12.0 in. 0.0419 ksi   2.16 kips  Vr2 3  23.1kips

n.g.

Rn 23  8.60 in.12.0 in. 0.0419 ksi    3.00   1.44 kips  Vr 23  15.5 kips n.g.

The calculations indicate that the bond strength is inadequate for all faces, thus an alternate means of load transfer such as the use of internal bearing plates as demonstrated previously in this example is necessary. As demonstrated by this example, direct bond interaction provides limited available strength for transfer of longitudinal shears and is generally only acceptable for lightly loaded columns or columns with low shear transfer requirements such as those with loads applied to both concrete fill and steel encasement simultaneously.

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EXAMPLE I.4 FILLED COMPOSITE MEMBER IN AXIAL COMPRESSION Given: Determine if the filled composite member illustrated in Figure I.4-1 is adequate for the indicated dead and live loads. Table IV-1B in Part IV will be used in this example. The composite member consists of an ASTM A500 Grade C HSS with normal weight (145 lb/ft3) concrete fill having a specified concrete compressive strength, f c = 5 ksi.

Fig. I.4-1. Filled composite member section and applied loading.

Solution: From AISC Manual Table 2-4, the material properties are: ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi From ASCE/SEI 7, Chapter 2, the required compressive strength is: LRFD

ASD

Pr  Pa

Pr  Pu  1.2  32 kips   1.6  84 kips 

 32 kips  84 kips

 173 kips

 116 kips

Method 1: AISC Tables The most direct method of calculating the available compressive strength is through the use of Table IV-1B (Part IV of this document). A K factor of 1.0 is used for a pin-ended member. Because the unbraced length is the same in both the x-x and y-y directions, and Ix exceeds Iy, y-y axis buckling will govern. Entering Table IV-1B with Lcy = KLy = 14 ft yields:

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LRFD

c Pn  368 kips  173 kips

ASD

Pn  245 kips  116 kips c

o.k.

o.k.

Method 2: AISC Specification Calculations As an alternate to using Table IV-1B, the available compressive strength can be calculated directly using the provisions of AISC Specification Chapter I. From AISC Manual Table 1-11 and Figure I.4-1, the geometric properties of an HSS106a are as follows: As H B tnom t h/t b/t Isx Isy

= 10.4 in.2 = 10.0 in. = 6.00 in. = a in. (nominal wall thickness) = 0.349 in. (design wall thickness) = 25.7 = 14.2 = 137 in.4 = 61.8 in.4

As shown in Figure I.1-1, internal clear distances are determined as: hi  H  2t  10.0 in.  2  0.349 in.  9.30 in. bi  B  2t  6.00 in.  2  0.349 in.  5.30 in.

From Design Example I.3, the area of concrete, Ac, equals 49.2 in.2 The steel and concrete areas can be used to calculate the gross cross-sectional area as follows:

Ag  As  Ac  10.4 in.2  49.2 in.2  59.6 in.2 Calculate the concrete moment of inertia using geometry compatible with that used in the calculation of the steel area in AISC Manual Table 1-11 (taking into account the design wall thickness and corner radii of two times the design wall thickness in accordance with AISC Manual Part 1), the following equations may be used, based on the terminology given in Figure I-1 in the introduction to these examples: For bending about the x-x axis:

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I cx 

 B  4t  hi3 12



t  H  4t 

3

6



 9

2



 64 t 4

36

 H  4t 4t   t 2    3   2

2





4





4

3 3 92  64  0.349 in.  6.00 in.  4  0.349 in.   9.30 in.  0.349 in. 10.0 in.  4  0.349 in.     12 6 36

4  0.349 in.  2 10.0 in.  4  0.349 in.    0.349 in.    2 3  

2

 353 in.4 For bending about the y-y axis: I cy 

 H  4t  bi3 12



t  B  4t  6

3

 9 

2



 64 t 4

36

 B  4t 4t   t 2    3   2

2

3 3 92  64  0.349 in. 10.0 in.  4  0.349 in.   5.30 in.  0.349 in. 6.00 in.  4  0.349 in.      12 6 36

 6.00 in.  4  0.349 in. 4  0.349 in.     0.349 in.    2 3  

2

2

 115 in.4 Limitations of AISC Specification Sections I1.3 and I2.2a (1)

3 ksi  f c  10 ksi

Concrete Strength: f c  5 ksi o.k.

(2) Specified minimum yield stress of structural steel:

Fy  75 ksi

Fy  50 ksi o.k. (3) Cross-sectional area of steel section:



10.4 in.2   0.01 59.6 in.2  0.596 in.2

As  0.01Ag



o.k.

There are no minimum longitudinal reinforcement requirements in the AISC Specification within filled composite members; therefore, the area of reinforcing bars, Asr, for this example is zero. Classify Section for Local Buckling In order to determine the strength of the composite section subject to axial compression, the member is first classified as compact, noncompact or slender in accordance with AISC Specification Table I1.1a.  p  2.26  2.26

E Fy 29, 000 ksi 50 ksi

 54.4

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h / t  25.7   controlling  max   b / t  14.2   25.7  controlling   p ; therefore, the section is compact

Available Compressive Strength The nominal axial compressive strength for compact sections without consideration of length effects, Pno, is determined from AISC Specification Section I2.2b as:

Pno  Pp

(Spec. Eq. I2-9a)

E   Pp  Fy As  C2 f c  Ac  Asr s  Ec  

(Spec. Eq. I2-9b)

where C2 = 0.85 for rectangular sections







Pno   50 ksi  10.4 in.2  0.85  5 ksi  49.2 in.2  0.0 in.2



 729 kips

Because the unbraced length is the same in both the x-x and y-y directions, the column will buckle about the weaker yy axis (the axis having the lower moment of inertia). Icy and Isy will therefore be used for calculation of length effects in accordance with AISC Specification Sections I2.2b and I2.1b as follows:  A  Asr  C3  0.45  3  s   0.9  Ag   10.4 in.2  0.0 in.2  0.45  3  59.6 in.2   0.973  0.9  0.9

Ec  wc1.5

(Spec. Eq. I2-13)    0.9 

f c



 145 lb/ft 3



1.5

5 ksi

 3,900 ksi

EI eff  Es I sy  Es I sr  C3 Ec I cy





(from Spec. Eq. I2-12)



  29, 000 ksi  61.8 in.  0 kip-in.  0.9  3,900 ksi  115 in. 4

2

4



2

 2, 200, 000 kip-in.

Pe  2  EI eff  /  Lc 

2

(Spec. Eq. I2-5)

where Lc = KL and K = 1.0 for a pin-ended member

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Pe 



2 2, 200, 000 kip-in.2



1.0 14 ft 12 in./ft    769 kips

2

Pno 729 kips  Pe 769 kips  0.948  2.25

Therefore, use AISC Specification Equation I2-2. Pno   Pn  Pno 0.658 Pe  

   

  729 kips  0.658 

(Spec. Eq. I2-2) 0.948

 490 kips

Check adequacy of the composite column for the required axial compressive strength: LRFD

ASD

c  0.75

 c  2.00

c Pn  0.75  490 kips 

Pn 490 kips  c 2.00  245 kips  116 kips o.k.

 368 kips  173 kips

o.k.

The values match those tabulated in Table IV-1B. Available Compressive Strength of Bare Steel Section Due to the differences in resistance and safety factors between composite and noncomposite column provisions, it is possible to calculate a lower available compressive strength for a composite column than one would calculate for the corresponding bare steel section. However, in accordance with AISC Specification Section I2.2b, the available compressive strength need not be less than that calculated for the bare steel member in accordance with Chapter E. From AISC Manual Table 4-3, for an HSS106a, KLy = 14 ft: LRFD

c Pn  331kips  368 kips

ASD

Pn  220 kips  245 kips c

Thus, the composite section strength controls and is adequate for the required axial compressive strength as previously demonstrated. Force Allocation and Load Transfer Load transfer calculations for external axial forces should be performed in accordance with AISC Specification Section I6. The specific application of the load transfer provisions is dependent upon the configuration and detailing of the connecting elements. Expanded treatment of the application of load transfer provisions is provided in Design Example I.3.

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EXAMPLE I.5 FILLED COMPOSITE MEMBER IN AXIAL TENSION Given: Determine if the filled composite member illustrated in Figure I.5-1 is adequate for the indicated dead load compression and wind load tension. The entire load is applied to the steel section.

Fig. I.5-1. Filled composite member section and applied loading. The composite member consists of an ASTM A500, Grade C, HSS with normal weight (145 lb/ft3) concrete fill having a specified concrete compressive strength, f c = 5 ksi.

Solution: From AISC Manual Table 2-4, the material properties are: ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi From AISC Manual Table 1-11, the geometric properties are as follows: HSS106a

As = 10.4 in.2

There are no minimum requirements for longitudinal reinforcement in the AISC Specification; therefore, it is common industry practice to use filled shapes without longitudinal reinforcement, thus Asr = 0. From ASCE/SEI 7, Chapter 2, the required compressive strength is (taking compression as negative and tension as positive): LRFD

ASD

Governing Uplift Load Combination  0.9 D  1.0W

Governing Uplift Load Combination  0.6 D  0.6W

Pr  Pu

Pr  Pa

 0.9  32 kips   1.0 100 kips 

 0.6  32 kips   0.6 100 kips 

 71.2 kips

 40.8 kips

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Available Tensile Strength Available tensile strength for a filled composite member is determined in accordance with AISC Specification Section I2.2c. Pn  As Fy  Asr Fysr





(Spec. Eq. I2-14)



 10.4 in.2  50 ksi   0 in.2

  60 ksi 

 520 kips LRFD

ASD

t  0.90

 t  1.67

t Pn  0.90  520 kips 

Pn 520 kips  t 1.67

 468 kips  71.2 kips

o.k.

 311 kips  40.8 kips

o.k.

For filled composite HSS members with no internal longitudinal reinforcing, the values for available tensile strength may also be taken directly from AISC Manual Table 5-4. The values calculated here match those for the limit state of yielding shown in Table 5-4. Force Allocation and Load Transfer Load transfer calculations are not required for filled composite members in axial tension that do not contain longitudinal reinforcement, such as the one under investigation, as only the steel section resists tension.

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EXAMPLE I.6 FILLED COMPOSITE MEMBER IN COMBINED AXIAL COMPRESSION, FLEXURE AND SHEAR Given: Using AISC design tables, determine if the filled composite member illustrated in Figure I.6-1 is adequate for the indicated axial forces, shears and moments that have been determined in accordance with the direct analysis method of AISC Specification Chapter C for the controlling ASCE/SEI 7 load combinations.

Fig. I.6-1. Filled composite member section and member forces. The composite member consists of an ASTM A500, Grade C, HSS with normal weight (145 lb/ft3) concrete fill having a specified concrete compressive strength, f c = 5 ksi.

Solution: From AISC Manual Table 2-4, the material properties are: ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi From AISC Manual Table 1-11 and Figure I.6-1, the geometric properties are as follows: HSS106a

H B tnom t h/t b/t As Isx Isy Zsx

= 10.0 in. = 6.00 in. = a in. (nominal wall thickness) = 0.349 in. (design wall thickness) = 25.7 = 14.2 = 10.4 in.2 = 137 in.4 = 61.8 in.4 = 33.8 in.3

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Additional geometric properties used for composite design are determined in Design Examples I.3 and I.4 as follows: hi = 9.30 in. bi = 5.30 in. Ac = 49.2 in.2 Ag = 59.6 in.2 Asr = 0 in.2 Ec = 3,900 ksi Icx = 353 in.4 Icy = 115 in.4

clear distance between HSS walls (longer side) clear distance between HSS walls (shorter side) cross-sectional area of concrete fill gross cross-sectional area of composite member area of longitudinal reinforcement modulus of elasticity of concrete moment of inertia of concrete fill about the x-x axis moment of inertia of concrete fill about the y-y axis

Limitations of AISC Specification Sections I1.3 and I2.2a (1)

3 ksi  f c  10 ksi

Concrete Strength: f c  5 ksi o.k.

(2) Specified minimum yield stress of structural steel:

Fy  75 ksi

Fy  50 ksi o.k. (3) Cross-sectional area of steel section:



10.4 in.   0.01 59.6 in. 2

2

2

 0.596 in.

As  0.01Ag



o.k.

Classify Section for Local Buckling The composite member in question was shown to be compact for pure compression in Example I.4 in accordance with AISC Specification Table I1.1a. The section must also be classified for local buckling due to flexure in accordance with Specification Table I1.1b; however, since the limits for members subject to flexure are equal to or less stringent than those for members subject to compression, the member is compact for flexure. Interaction of Axial Force and Flexure The interaction between axial forces and flexure in composite members is governed by AISC Specification Section I5 which, for compact members, permits the use of the methods of Section I1.2 with the option to use the interaction equations of Section H1.1. The strain compatibility method is a generalized approach that allows for the construction of an interaction diagram based upon the same concepts used for reinforced concrete design. Application of the strain compatibility method is required for irregular/nonsymmetrical sections, and its general application may be found in reinforced concrete design texts and will not be discussed further here. Plastic stress distribution methods are discussed in AISC Specification Commentary Section I5 which provides three acceptable procedures for compact filled members. The first procedure, Method 1, invokes the interaction equations of Section H1. The second procedure, Method 2, involves the construction of a piecewise-linear interaction curve using the plastic strength equations provided in AISC Manual Table 6-4. The third procedure, Method 2— Simplified, is a reduction of the piecewise-linear interaction curve that allows for the use of less conservative interaction equations than those presented in Chapter H (refer to AISC Specification Commentary Figure C-I5.3). For this design example, each of the three applicable plastic stress distribution procedures are reviewed and compared. Method 1: Interaction Equations of Section H1

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The most direct and conservative method of assessing interaction effects is through the use of the interaction equations of AISC Specification Section H1. For HSS shapes, both the available compressive and flexural strengths can be determined from Table IV-1B (included in Part IV of this document). In accordance with the direct analysis method, a K factor of 1 is used. Because the unbraced length is the same in both the x-x and y-y directions, and Ix exceeds Iy, y-y axis buckling will govern for the compressive strength. Flexural strength is determined for the x-x axis to resist the applied moment about this axis indicated in Figure I.6-1. Entering Table IV-1B with Lcy = 14 ft yields: LRFD

ASD

c Pn  368 kips b M nx  141 kip-ft

Pn  c  245 kips M nx  b  93.5 kip-ft

Pr P  u Pc c Pn 129 kips  368 kips

Pr Pa  Pc Pn /  c 98.2 kips  245 kips  0.401  0.2

 0.351  0.2

Therefore, use AISC Specification Equation H1-1a.

Therefore, use AISC Specification Equation H1-1a.

Pu 8  Mu   c Pn 9  b M n

Pa 8  Ma  Pn / c 9  M n / b

   1.0 

(from Spec. Eq. H1-1a)

   1.0 

(from Spec. Eq. H1-1a)

129 kips 8  120 kip-ft      1.0 368 kips 9  141 kip-ft 

98.2 kips 8  54 kip-ft      1.0 245 kips 9  93.5 kip-ft 

1.11  1.0

0.914  1.0

n.g.

o.k.

Using LRFD methodology, Method 1 indicates that the section is inadequate for the applied loads. The designer can elect to choose a new section that passes the interaction check or re-analyze the current section using a less conservative design method such as Method 2. The use of Method 2 is illustrated in the following section. Using ASD methodology, Method 1 indicates that the section is adequate for the applied loads. Method 2: Interaction Curves from the Plastic Stress Distribution Model The procedure for creating an interaction curve using the plastic stress distribution model is illustrated graphically in Figure I.6-2. Referencing Figure I.6-2, the nominal strength interaction surface A, B, C, D, E is first determined using the equations provided in AISC Manual Table 6-4. This curve is representative of the short column member strength without consideration of length effects. A slenderness reduction factor, , is then calculated and applied to each point to create surface A , B, C, D , E . The appropriate resistance or safety factors are then applied to create the design surface A , B, C, D , E . Finally, the required axial and flexural strengths from the applicable load combinations of ASCE/SEI 7 are plotted on the design surface, and the member is acceptable for the applied loading if all points fall within the design surface. These steps are illustrated in detail by the following calculations. Step 1: Construct nominal strength interaction surface A, B, C, D, E without length effects Using the equations provided in AISC Manual Table 6-4 for bending about the x-x axis yields: Point A (pure axial compression):

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PA  Fy As  0.85 f cAc







  50 ksi  10.4 in.2  0.85  5 ksi  49.2 in.2



 729 kips M A  0 kip-ft

Point D (maximum nominal moment strength):

PD  

0.85 f cAc 2



0.85  5 ksi  49.2 in.2



2

 105 kips

Z sx  33.8 in.3 ri  t  0.349 in. Zc  

bi hi2  0.429ri 2 hi  0.192ri 3 4

 5.30 in. 9.30 in.2 4

 0.429  0.349 in.  9.30 in.  0.192  0.349 in. 2

3

 114 in.3

Fig. I.6-2. Interaction diagram for composite beam-column—Method 2.

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M D  Fy Z sx 

0.85 f cZ c 2



 0.85  5 ksi  114 in.3   50 ksi  33.8 in.3   2   161 kip-ft





  

1    12 in./ft  

Point B (pure flexure): PB  0 kips

hn  

0.85 f cAc h  i 2  0.85 f cbi  4 Fy t  2



0.85  5 ksi  49.2 in.2



2 0.85  5 ksi  5.30 in.  4  50 ksi  0.349 in. 



9.30 in. 2

 1.13 in.  4.65 in.  1.13 in. Z sn  2thn2  2  0.349 in.1.13 in.

2

 0.891 in.3

Z cn  bi hn2   5.30 in.1.13 in.

2

 6.77 in.3 Z  M B  M D  Fy Z sn  0.85 f c  cn   2   6.77 in.3      1 1  161 kip-ft   50 ksi  0.891 in.3      0.85  5 ksi    12 in./ft 2 12 in./ft       156 kip-ft





Point C (intermediate point): PC  0.85 f cAc



 0.85  5 ksi  49.2 in.2



 209 kips MC  M B  156 kip-ft

Point E (optional): Point E is an optional point that helps better define the interaction curve.

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hn H where hn  1.13 in. from Point B  2 4 1.13 in. 10.0 in.   2 4  3.07 in.

hE 

PE  

0.85 f cAc  0.85 f cbi hE  4 Fy thE 2



0.85  5 ksi  49.2 in.2 2

  0.85  5 ksi 5.30 in. 3.07 in.  4 50 ksi  0.349 in.3.07 in.

 388 kips Z cE  bi hE2   5.30 in. 3.07 in.

2

 50.0 in.3 Z sE  2thE2  2  0.349 in. 3.07 in.

2

 6.58 in.3

M E  M D  Fy Z sE 

0.85 f cZ cE 2



3   1   0.85  5 ksi  50.0 in.  161 kip-ft   50 ksi  6.58 in.3    2  12 in./ft     125 kip-ft





  

1    12 in./ft  

The calculated points are plotted to construct the nominal strength interaction surface without length effects as depicted in Figure I.6-3. Step 2: Construct nominal strength interaction surface A , B, C, D , E  with length effects The slenderness reduction factor, , is calculated for Point A using AISC Specification Section I2.2 in accordance with Specification Commentary Section I5.

Pno  PA  729 kips  A  Asr  C3  0.45  3  s   0.9  Ag   10.4 in.2  0 in.2  0.45  3  59.6 in.2   0.973  0.9  0.9

(Spec. Eq. I2-13)    0.9 

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EI eff  Es I sy  Es I sr  C3 Ec I cy





(from Spec. Eq. I2-12)



  29, 000 ksi  61.8 in.4  0  0.9  3,900 ksi  115 in.4



 2, 200, 000 ksi Pe  2  EI eff 

  Lc 2 , where Lc  KL and K  1.0 in accordance with the direct analysis method

(Spec. Eq. I2-5)

 2, 200, 000 ksi  2 14 ft 12 in./ft   

2

 769 kips

Pno 729 kips  Pe 769 kips  0.948  2.25 Use AISC Specification Equation I2-2. Pno  Pn  Pno  0.658 Pe  

   

(Spec. Eq. I2-2)

  729 kips  0.658 

0.948

 490 kips

Fig. I.6-3. Nominal strength interaction surface without length effects.

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From AISC Specification Commentary Section I5: Pn Pno 490 kips  729 kips



 0.672

In accordance with AISC Specification Commentary Section I5, the same slenderness reduction is applied to each of the remaining points on the interaction surface as follows: PA  PA  0.672  729 kips   490 kips PB  PB  0.672  0 kip   0 kip PC   PC  0.672  209 kips   140 kips PD  PD  0.672 105 kips   70.6 kips PE   PE  0.672  388 kips   261 kips

The modified axial strength values are plotted with the flexural strength values previously calculated to construct the nominal strength interaction surface including length effects. These values are superimposed on the nominal strength surface not including length effects for comparison purposes in Figure I.6-4. Step 3: Construct design interaction surface A , B, C, D , E  and verify member adequacy The final step in the Method 2 procedure is to reduce the interaction surface for design using the appropriate resistance or safety factors. LRFD

ASD

Design compressive strength: c  0.75

Allowable compressive strength:  c  2.00

PX   c PX  , where X  A, B, C, D or E

PX   PX  /  c , where X  A, B, C, D or E

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LRFD

PA  0.75  490 kips 

PA  490 kips 2.00  245 kips

 368 kips PB  0.75  0 kip 

PB  0 kip 2.00  0 kip

 0 kip PC   0.75 140 kips 

PC   140 kips 2.00  70.0 kips

 105 kips PD  0.75  70.6 kips   53.0 kips

PD  70.6 kips 2.00  35.3 kips

PE   0.75  261 kips   196 kips

ASD

PE   261 kips 2.00  131 kips

Fig. I.6-4. Nominal strength interaction surfaces (with and without length effects).

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LRFD

ASD

Design flexural strength: b  0.90

Allowable flexural strength:  b  1.67

M X   b M X , where X = A, B, C, D or E

M X   M X  b , where X = A, B, C, D or E

M A  0.90  0 kip-ft 

M A  0 kip-ft 1.67

 0 kip-ft M B  0.90 156 kip-ft 

 0 kip-ft M B  156 kip-ft 1.67

 140 kip-ft

 93.4 kip-ft

M C   0.90 156 kip-ft 

M C   156 kip-ft 1.67

 140 kip-ft

 93.4 kip-ft

M D  0.90 161 kip-ft 

M D  161 kip-ft 1.67

 145 kip-ft

 96.4 kip-ft

M E   0.90 124 kip-ft 

M E   124 kip-ft 1.67

 112 kip-ft

 74.3 kip-ft

The available strength values for each design method can now be plotted. These values are superimposed on the nominal strength surfaces (with and without length effects) previously calculated for comparison purposes in Figure I.6-5. By plotting the required axial and flexural strength values determined for the governing load combinations on the available strength surfaces indicated in Figure I.6-5, it can be seen that both ASD (Ma, Pa) and LRFD (Mu, Pu) points lie within their respective design surfaces. The member in question is therefore adequate for the applied loads. Designers should carefully review the proximity of the available strength values in relation to point D on Figure I.65 as it is possible for point D to fall outside of the nominal strength curve, thus resulting in an unsafe design. This possibility is discussed further in AISC Specification Commentary Section I5 and is avoided through the use of Method 2—Simplified as illustrated in the following section. Method 2: Simplified The simplified version of Method 2 involves the removal of points D and E  from the Method 2 interaction surface leaving only points A , B and C as illustrated in the comparison of the two methods in Figure I.6-6. Reducing the number of interaction points allows for a bilinear interaction check defined by AISC Specification Commentary Equations C-I5-1a and C-I5-1b to be performed. Using the available strength values previously calculated in conjunction with the Commentary equations, interaction ratios are determined as follows:

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LRFD

ASD

Pr  Pu  129 kips

Pr  Pa  98.2 kips

Pr  PC   105 kips

Pr  PC   70.0 kips

Therefore, use AISC Specification Commentary Equation C-I5-1b.

Therefore, use AISC Specification Commentary Equation C-I5-1b.

Pr  PC M r   1.0 PA  PC M C

(from Spec. Eq. C-I5-1b)

Pr  PC M r   1.0 PA  PC M C

(from Spec. Eq. C-I5-1b)

which for LRFD equals:

which for ASD equals:

Pu  PC  M  u  1.0 PA  PC  M C  129 kips  105 kips 120 kip-ft   1.0 368 kips  105 kips 140 kip-ft

Pa  PC  M  a  1.0 PA  PC  M C  98.2 kips  70.0 kips 54 kip-ft   1.0 245 kips  70.0 kips 93.4 kip-ft

0.948  1.0

0.739  1.0

o.k.

o.k.

Thus, the member is adequate for the applied loads.

Fig. I.6-5. Available and nominal interaction surfaces.

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Comparison of Methods The composite member was found to be inadequate using Method 1—Chapter H interaction equations, but was found to be adequate using both Method 2 and Method 2—Simplified procedures. A comparison between the methods is most easily made by overlaying the design curves from each method as illustrated in Figure I.6-7 for LRFD design. From Figure I.6-7, the conservative nature of the Chapter H interaction equations can be seen. Method 2 provides the highest available strength; however, the Method 2—Simplified procedure also provides a good representation of the complete design curve. By using the Part IV design tables to determine the available strength of the composite member in compression and flexure (Points A  and B respectively), the modest additional effort required to calculate the available compressive strength at Point C  can result in appreciable gains in member strength when using Method 2—Simplified as opposed to Method 1.

Fig. I.6-6. Comparison of Method 2 and Method 2—Simplified.

Fig. I.6-7. Comparison of interaction methods (LRFD).

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Available Shear Strength AISC Specification Section I4.1 provides three methods for determining the available shear strength of a filled composite member: available shear strength of the steel section alone in accordance with Chapter G; available shear strength of the reinforced concrete portion alone per ACI 318 (ACI 318, 2014); or available shear strength of the steel section plus the reinforcing steel ignoring the contribution of the concrete. The available shear strength will be determined using the first two methods because there is no reinforcing steel provided in this example. Available Shear Strength of Steel Section The nominal shear strength, Vn, of rectangular HSS members is determined using the provisions of AISC Specification Section G4. The web shear coefficient, Cv2, is determined from AISC Specification Section G2.2 with, h/tw = h/t and kv = 5.

1.10 kv E Fy  1.10

 5 29, 000 ksi  50 ksi

 59.2  h t  25.7 Use AISC Specification Equation G2-9. Cv 2  1.0

(Spec. Eq. G2-9)

The nominal shear strength is calculated as: h  H  3t

 10.0 in.  3  0.349 in.  8.95 in.

Aw  2ht  2  8.95 in. 0.349 in.  6.25 in.2 Vn  0.6 Fy AwCv 2



(Spec. Eq. G4-1)



 0.6  50 ksi  6.25 in.2 1.0   188 kips

The available shear strength of the steel section is: LRFD

ASD

v  0.90

 v  1.67

vVn  0.90 188 kips 

Vn 188 kips  v 1.67  113 kips  10.3 kips o.k.

 169 kips  17.1 kips o.k.

Available Shear Strength of the Reinforced Concrete The available shear strength of the steel section alone has been shown to be sufficient, but the available shear strength of the concrete will be calculated for demonstration purposes. Considering that the member does not have

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longitudinal reinforcing, the method of shear strength calculation involving reinforced concrete is not valid; however, the design shear strength of the plain concrete using ACI 318, Chapter 14, can be determined as follows:  = 0.60 for plain concrete design from ACI 318 Section 21.2.1  = 1.0 for normal weight concrete from ACI 318 Section 19.2.4.2 4 Vn     f cbw h 3

(ACI 318 Section 14.5.5.1)

bw  bi h  hi

 1 kip  4 Vn    1.0  5, 000 psi  5.30 in. 9.30 in.   3    1, 000 lb   4.65 kips

Vn  0.60  4.65 kips   2.79 kips  17.1 kips

(ACI 318 Section 14.5.1.1) n.g.

As can be seen from this calculation, the shear resistance provided by plain concrete is small and the strength of the steel section alone is generally sufficient. Force Allocation and Load Transfer Load transfer calculations for applied axial forces should be performed in accordance with AISC Specification Section I6. The specific application of the load transfer provisions is dependent upon the configuration and detailing of the connecting elements. Expanded treatment of the application of load transfer provisions is provided in Design Example I.3.

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EXAMPLE I.7 FILLED COMPOSITE BOX COLUMN WITH NONCOMPACT/SLENDER ELEMENTS Given:

Determine the required ASTM A36 plate thickness of the filled composite box column illustrated in Figure I.7-1 to resist the indicated axial forces, shears and moments that have been determined in accordance with the direct analysis method of AISC Specification Chapter C for the controlling ASCE/SEI 7 load combinations. The core is composed of normal weight (145 lb/ft3) concrete fill having a specified concrete compressive strength, f c = 7 ksi.

Fig. I.7-1. Composite box column section and member forces. Solution:

From AISC Manual Table 2-5, the material properties are: ASTM A36 Fy = 36 ksi Fu = 58 ksi Trial Size 1 (Noncompact)

For ease of calculation the contribution of the plate extensions to the member strength will be ignored as illustrated by the analytical model in Figure I.7-1. Trial Plate Thickness and Geometric Section Properties of the Composite Member Select a trial plate thickness, t, of a in. Note that the design wall thickness reduction of AISC Specification Section B4.2 applies only to electric-resistance-welded HSS members and does not apply to built-up sections such as the one under consideration. The calculated geometric properties of the 30 in. by 30 in. steel box column are:

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B  30 in. H  30 in. Ag  900 in.2 Ac  856 in.2 As  44.4 in.2 bi  B  2t  30 in.  2  a in.  29.2 in. hi  H  2t  30 in.  2  a in.  29.2 in.

Ec  wc1.5 f c



 145 lb/ft 3



1.5

7 ksi

 4, 620 ksi I gx  

BH 3 12

 30 in. 30 in.3

12  67,500 in.4 I cx  

bi hi 3 12

 29.2 in. 29.2 in.3

12  60, 600 in.4

I sx  I gx  I cx  67,500 in.4  60, 600 in.4  6,900 in.4 Limitations of AISC Specification Sections I1.3 and I2.2a (1) Concrete Strength: f c  7 ksi o.k.

3 ksi  f c  10 ksi

(2) Specified minimum yield stress of structural steel:

Fy  75 ksi

Fy  36 ksi o.k.

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(3) Cross-sectional area of steel section:



44.4 in.2   0.01 900 in.2  9.00 in.2

As  0.01Ag



o.k.

Classify Section for Local Buckling Classification of the section for local buckling is performed in accordance with AISC Specification Table I1.1a for compression and Table I1.1b for flexure. As noted in Specification Section I1.4, the definitions of width, depth and thickness used in the evaluation of slenderness are provided in Section B4.1b. For box columns, the widths of the stiffened compression elements used for slenderness checks, b and h, are equal to the clear distances between the column walls, bi and hi. The slenderness ratios are determined as follows: bi hi  t t 29.2 in.  a in.



 77.9

Classify section for local buckling in steel elements subject to axial compression from AISC Specification Table I1.1a:  p  2.26  2.26

E Fy 29, 000 ksi 36 ksi

 64.1  r  3.00  3.00

E Fy 29, 000 ksi 36 ksi

 85.1  p     r ; therefore, the section is noncompact for compression

According to AISC Specification Section I1.4, if any side of the section in question is noncompact or slender, then the entire section is treated as noncompact or slender. For the square section under investigation; however, this distinction is unnecessary as all sides are equal in length. Classification of the section for local buckling in elements subject to flexure is performed in accordance with AISC Specification Table I1.1b. Note that flanges and webs are treated separately; however, for the case of a square section only the most stringent limitations, those of the flange, need be applied. Noting that the flange limitations for bending are the same as those for compression,  p     r ; therefore, the section is noncompact for flexure

Available Compressive Strength

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Compressive strength for noncompact filled composite members is determined in accordance with AISC Specification Section I2.2b(b). E   Pp  Fy As  C2 f c  Ac  Asr s  , where C2  0.85 for rectangular sections Ec  







  36 ksi  44.4 in.2  0.85  7 ksi  856 in.2  0 in.2

(Spec. Eq. I2-9b)



 6, 690 kips E   Py  Fy As  0.7 f c  Ac  Asr s  E c  





(Spec. Eq. I2-9d)



  36 ksi  44.4 in.2  0.7  7 ksi  856 in.2  0 in.2



 5, 790 kips Pno  Pp 

Pp  Py

 r   p 

 6, 690 kips 

2

   p 

2

(Spec. Eq. I2-9c)

6, 690 kips  5, 790 kips

85.1  64.1

2

 77.9  64.12

 6,300 kips  A  Asr  C3  0.45  3  s   0.9  Ag   44.4 in.2  0 in.2  0.45  3  900 in.2   0.598  0.9 = 0.598 EI eff  Es I s  Es I sr  C3 Ec I c



(Spec. Eq. I2-13)    0.9 

(Spec. Eq. I2-12)





  29, 000 ksi  6,900 in.  0.0 kip-in.  0.598  4, 620 ksi  60, 600 in. 4

2

4



 368, 000, 000 kip-in.2

Pe  2  EI eff  /  Lc  , where Lc  KL and K =1.0 in accordance with the direct analysis method 2





2 368, 000, 000 kip-in.2

 30 ft 12 in./ft    28, 000 kips



2

Pno 6,300 kips  Pe 28, 000 kips  0.225  2.25

Therefore, use AISC Specification Equation I2-2.

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Pno  Pn  Pno  0.658 Pe  

   

(Spec. Eq. I2-2)

  6,300 kips  0.658 

0.225

 5, 730 kips

According to AISC Specification Section I2.2b, the compression strength need not be less than that specified for the bare steel member as determined by Specification Chapter E. It can be shown that the compression strength of the bare steel for this section is equal to 955 kips, thus the strength of the composite section controls. The available compressive strength is: LRFD

ASD

c  0.75

 c  2.00

c Pn  0.75  5, 730 kips 

Pn 5, 730 kips  c 2.00  2,870 kips

 4,300 kips

Available Flexural Strength Flexural strength of noncompact filled composite members is determined in accordance with AISC Specification Section I3.4b(b): Mn  M p  M p  M y 

   p  r   p 

(Spec. Eq. I3-3b)

In order to utilize Equation I3-3b, both the plastic moment strength of the section, Mp, and the yield moment strength of the section, My, must be calculated. Plastic Moment Strength The first step in determining the available flexural strength of a noncompact section is to calculate the moment corresponding to the plastic stress distribution over the composite cross section, Mp. This concept is illustrated graphically in AISC Specification Commentary Figure C-I3.7(a) and follows the force distribution depicted in Figure I.7-2 and detailed in Table I.7-1.

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Table I.7-1. Plastic Moment Equations Component

Force

Compression in steel flange

C1  bi tf Fy

Compression in concrete

C2  0.85fc  a p  t f  bi

Compression in steel web

C3  ap 2tw Fy

Tension in steel web

T1   H  ap  2tw Fy

Tension in steel flange

T2  bi tf Fy

Moment Arm t y C1  ap  f 2 ap  tf yC 2  2 ap yC 3  2 H  ap yT 1  2 yT 2  H  a p 

tf 2

where: ap  Mp 

2Fy Htw  0.85fcbi tf 4tw Fy  0.85fcbi

  force moment arm

Using the equations provided in Table I.7-1 for the section in question results in the following:

ap 

2  36 ksi  30 in. a in.  0.85  7 ksi  29.2 in. a in. 4  a in. 36 ksi   0.85  7 ksi  29.2 in.

 3.84 in.

Figure I.7-2. Plastic moment stress blocks and force distribution.

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Force C1   29.2 in. a in. 36 ksi   394 kips

C2  0.85  7 ksi  3.84 in.  a in. 29.2 in.  602 kips C3   3.84 in. 2  a in. 36 ksi 

T1   30 in.  3.84 in. 2  a in. 36 ksi 

T2   29.2 in. a in. 36 ksi   394 kips

Mp 

C1yC1  1, 440 kip-in.

3.84 in.  a in. 2  1.73 in.

C2 yC 2  1,040 kip-in.

3.84 in. 2  1.92 in.

C3 y C 3  200 kip-in.

30 in.  3.84 in. 2  13.1 in.

T1yT 1  9,250 kip-in.

yC 2 

yC 3 

 104 kips

 706 kips

Force  Moment Arm

Moment Arm a in. yC1  3.84 in.  2  3.65 in.

yT 1 

yT 2  30 in.  3.84 in. 

a in. 2

 26.0 in.

T2 yT 2  10,200 kip-in.

  force moment arm

1,440 kip-in.  1,040 kip-in.  200 kip-in.  9,250 kip-in.  10,200 kip-in.  12 in./ft  1,840 kip-ft

Yield Moment Strength The next step in determining the available flexural strength of a noncompact filled member is to determine the yield moment strength. The yield moment is defined in AISC Specification Section I3.4b(b) as the moment corresponding to first yield of the compression flange calculated using a linear elastic stress distribution with a maximum concrete compressive stress of 0.7 f c . This concept is illustrated diagrammatically in Specification Commentary Figure C-I3.7(b) and follows the force distribution depicted in Figure I.7-3 and detailed in Table I.7-2.

Figure I.7-3. Yield moment stress blocks and force distribution.

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Table I.7-2. Yield Moment Equations Component

Force

Moment Arm

Compression in steel flange

C1  bi tf Fy

t y C 1  ay  f 2

Compression in concrete

C2  0.35fc  ay  tf  bi

yC 2 

Compression in steel web

C3  ay 2tw 0.5Fy

T2   H  2ay  2tw Fy

Tension in steel flange

3

2ay yC 3  3 2ay yT 1  3 H yT 2  2

T1  ay 2tw 0.5Fy

Tension in steel web

2  ay  tf 

T3  bi tf Fy

yT 3  H  a y 

tf 2

where ay  My 

2Fy Htw  0.35fcbi tf 4tw Fy  0.35fcbi

  force moment arm 

Using the equations provided in Table I.7-2 for the section in question results in the following:

ay 

2  36 ksi  30 in. a in.  0.35  7 ksi  29.2 in. a in. 4  a in. 36 ksi   0.35  7 ksi  29.2 in.

 6.66 in. Force C1   29.2 in. a in. 36 ksi   394 kips C2  0.35  7 ksi  6.66 in.  a in. 29.2 in.  450 kips

yC 2 

 89.9 kips T1   6.66 in. 2  a in. 0.5  36 ksi   89.9 kips

2  6.66 in.  a in. C2 yC 2  1,890 kip-in.

3

yC 3 

2  6.66 in. C3 y C 3  399 kip-in.

3  4.44 in.

yT 1 

2  6.66 in. T1yT 1  399 kip-in.

3  4.44 in. 30 in. 2  15.0 in.

T2  30 in.  2  6.66 in.   2  a in. 36 ksi  450 kips

yT 2 

T3   29.2 in. a in. 36 ksi 

yT 3  30 in.  6.66 in. 

My 

C1y C1  2,550 kip-in.

 4.19 in.

C3   6.66 in. 2  a in. 0.5  36 ksi 

 394 kips

Force  Moment Arm

Moment Arm a in. yC1  6.66 in.  2  6.47 in.

T2 yT 2  6,750 kip-in.

a in. 2

 23.2 in.

T3 yT 3  9,140 kip-in.

  force moment arm

2,550 kip-in.  1,890 kip-in.  399 kip-in.  399 kip-in.  6,750 kip-in.  9,140 kip-in. 12 in./ft  1,760 kip-ft 

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Now that both Mp and My have been determined, Equation I3-3b may be used in conjunction with the flexural slenderness values previously calculated to determine the nominal flexural strength of the composite section as follows:

  p M n  M p   M p  M y    r   p

  

(Spec. Eq. I3-3b)

 77.9  64.1   1,840 kip-ft  1,840 kip-ft  1, 760 kip-ft     85.1  64.1   1, 790 kip-ft The available flexural strength is: LRFD

ASD

b  0.90

 b  1.67

b M n  0.90 1, 790 kip-ft 

M n 1, 790 kip-ft  b 1.67  1, 070 kip-ft

 1, 610 kip-ft

Interaction of Flexure and Compression Design of members for combined forces is performed in accordance with AISC Specification Section I5. For filled composite members with noncompact or slender sections, interaction may be determined in accordance with Section H1.1 as follows: LRFD

ASD

Pu  1,310 kips M u  552 kip-ft

Pa  1,370 kips M a  248 kip-ft

Pr P  u Pc c Pn 1,310 kips  4,300 kips  0.305  0.2

Pr Pa  Pc Pn / c 1,370 kips  2,870 kips  0.477  0.2

Therefore, use AISC Specification Equation H1-1a.

Therefore, use AISC Specification Equation H1-1a.

Pu 8  Mu    (from Spec. Eq. H1-1a)   1.0 c Pn 9  b M n  8  552 kip-ft  0.305     1.0 9  1, 610 kip-ft 

Pa 8  Ma   Pn / c 9  M n / b

0.610  1.0

o.k.

   1.0 

(from Spec. Eq. H1-1a)

8  248 kip-ft  0.477     1.0 9  1, 070 kip-ft  0.683  1.0 o.k.

The composite section is adequate; however, as there is available strength remaining for the trial plate thickness chosen, re-analyze the section to determine the adequacy of a reduced plate thickness. Trial Size 2 (Slender)

The calculated geometric section properties using a reduced plate thickness of t = 4 in. are:

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B  30 in. H  30 in. Ag  900 in.2 Ac  870 in.2 As  29.8 in.2 bi  B  2t  30 in.  2 4 in.  29.5 in. hi  H  2t  30 in.  2 4 in.  29.5 in.

Ec  wc1.5 f c



 145 lb/ft 3



1.5

7 ksi

 4, 620 ksi I gx  

BH 3 12

 30 in. 30 in.3

12  67,500 in.4 I cx  

bi hi 3 12

 29.5 in. 29.5 in.3

12  63,100 in.4

I sx  I gx  I cx  67,500 in.4  63,100 in.4  4, 400in.4 Limitations of AISC Specification Sections I1.3 and I2.2a (1) Concrete Strength: f c  7 ksi o.k.

3 ksi  f c  10 ksi

(2) Specified minimum yield stress of structural steel:

Fy  75 ksi

Fy  36 ksi o.k.

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(3) Cross sectional area of steel section:



29.8 in.2   0.01 900 in.2  9.00 in.2

As  0.01Ag



o.k.

Classify Section for Local Buckling As noted previously, the definitions of width, depth and thickness used in the evaluation of slenderness are provided in AISC Specification Section B4.1b. For a box column, the slenderness ratio is determined as the ratio of clear distance-to-wall thickness: bi hi  t t 29.5 in.  4 in.



 118

Classify section for local buckling in steel elements subject to axial compression from AISC Specification Table I1.1a. As determined previously, r = 85.1.  max  5.00  5.00

E Fy 29, 000 ksi 36 ksi

 142  r     max ; therefore, the section is slender for compression

Classification of the section for local buckling in elements subject to flexure occurs separately per AISC Specification Table I1.1b. Because the flange limitations for bending are the same as those for compression,  r     max ; therefore, the section is slender for flexure

Available Compressive Strength Compressive strength for a slender filled member is determined in accordance with AISC Specification Section I2.2b(c). Fcr 



9 Es

(Spec. Eq. I2-10)

2

b   t 9  29, 000 ksi 

1182

 18.7 ksi

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E   Pno  Fcr As  0.7 f c  Ac  Asr s  Ec  



(Spec. Eq. I2-9e)





 18.7 ksi  29.8 in.2  0.7  7 ksi  870 in.2  0 in.2



 4,820 kips  A  Asr  C3  0.45  3  s   0.9  Ag   29.8 in.2  0 in.2   0.45  3    0.9 900 in.2    0.549  0.9  0.549 EI eff  Es I s  Es I sr  C3 Ec I c

(Spec. Eq. I2-13)

(Spec. Eq. I2-12)







  29, 000 ksi  4, 400 in.4  0 kip-in.2  0.549  4, 620 ksi  63,100 in.4



 288, 000, 000 kip-in.2

Pe  2  EI eff  /  Lc  , where Lc  KL and K  1.0 in accordance with the direct analysis method (Spec. Eq. I2-5) 2





2 288, 000, 000 kip-in.2

 30 ft 12 in./ft    21,900 kips



2

Pno 4,820 kips  Pe 21,900 kips  0.220  2.25

Therefore, use AISC Specification Equation I2-2. Pno   Pn  Pno 0.658 Pe  

   

  4,820 kips  0.658 

(Spec. Eq. I2-2) 0.220

 4, 400 kips

According to AISC Specification Section I2.2b the compression strength need not be less than that determined for the bare steel member using Specification Chapter E. It can be shown that the compression strength of the bare steel for this section is equal to 450 kips, thus the strength of the composite section controls. The available compressive strength is:

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LRFD

ASD

c  0.75

 c  2.00

c Pn  0.75  4, 400 kips 

Pn 4, 400 kips  c 2.00  2, 200 kips

 3,300 kips

Available Flexural Strength Flexural strength of slender filled composite members is determined in accordance with AISC Specification Section I3.4b(c). The nominal flexural strength is determined as the first yield moment, Mcr, corresponding to a flange compression stress of Fcr using a linear elastic stress distribution with a maximum concrete compressive stress of 0.7 f c . This concept is illustrated diagrammatically in Specification Commentary Figure C-I3.7(c) and follows the force distribution depicted in Figure I.7-4 and detailed in Table I.7-3.

Table I.7-3. First Yield Moment Equations Component

Force

Moment Arm

Compression in steel flange

C1  bi tf Fcr

yC1  acr

Compression in concrete

C2  0.35fc  acr  tf  bi

yC 2 

t  f 2

2  acr  tf  3

2a  cr 3

Compression in steel web

C3  acr 2tw 0.5Fcr

yC 3

Tension in steel web

T1   H  acr  2tw 0.5Fy

yT 1 

Tension in steel flange

T2  bi tf Fy

yT 2  H  acr 

where:

acr  Mcr 

2  H  acr  3 tf 2

Fy Htw   0.35fc  Fy  Fcr  bi tf tw  Fcr  Fy   0.35fc bi

  force moment arm

Using the equations provided in Table I.7-3 for the section in question results in the following: acr 

 36 ksi  30 in.4 in.  0.35  7 ksi   36 ksi  18.7 ksi   29.5 in.4 in. 4 in.18.7 ksi  36 ksi   0.35  7 ksi  29.5 in.

 4.84 in.

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Force C1   29.5 in. 4 in.18.7 ksi   138 kips C2  0.35  7 ksi  4.84 in.  4 in. 29.5 in.  332 kips

yC 2 

C1yC1  651 kip-in.

2  4.84 in.  4 in. C2 yC 2  1,020 kip-in.

3

 3.06 in.

C3   4.84 in. 2  4 in. 0.5 18.7 ksi   22.6 kips T1   30 in.  4.84 in. 2  4 in. 0.5  36 ksi   226 kips

yC 3 

2  4.84 in. C3 yC 3  73.0 kip-in.

3  3.23 in.

yT 1 

2  30 in.  4.84 in. T1yT 1  3,800 kip-in.

3

 16.8 in.

T2   29.5 in. 4 in. 36 ksi 

yT 2  30 in.  4.84 in. 

 266 kips

Mcr 

Force  Moment Arm

Moment Arm 4 in. yC1  4.84 in.  2  4.72 in.

4 in. 2

 25.0 in.

T2 yT 2  6,650 kip-in.

  force component moment arm

651 kip-in.  1,020 kip-in.  73.0 kip-in.  3,800 kip-in.  6,650 kip-in. 12 in./ft  1,020 kip-ft 

The available flexural strength is: LRFD

ASD

b  0.90

 b  1.67

M n  0.90 1, 020 kip-ft 

M n 1, 020 kip-ft  b 1.67  611 kip-ft

 918 kip-ft

Figure I.7-4. First yield moment stress blocks and force distribution.

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Interaction of Flexure and Compression The interaction of flexure and compression may be determined in accordance with AISC Specification Section H1.1 as follows: LRFD

ASD

Pu  1,310 kips M u  552 kip-ft

Pa  1,370 kips M a  248 kip-ft

Pr P  u Pc c Pn 1,310 kips  3,300 kips

Pr Pa  Pc Pn /  c 1,370 kips  2, 200 kips  0.622  0.2

 0.397  0.2

Therefore, use AISC Specification Equation H1-1a.

Therefore, use AISC Specification Equation H1-1a.

Pu 8  Mu      1.0 c Pn 9  b M n  8  552 kip-ft  0.397     1.0 9  918 kip-ft 

Pa 8  Ma      1.0 Pn / c 9  M n / c 

0.931  1.0

(from Spec. Eq. H1-1a)

(from Spec. Eq. H1-1a)

8  248 kip-ft  0.622     1.0 9  611 kip-ft  0.983  1.0 o.k.

o.k.

Thus, a plate thickness of 4 in. is adequate. Note that in addition to the design checks performed for the composite condition, design checks for other load stages should be performed as required by AISC Specification Section I1. These checks should take into account the effect of hydrostatic loads from concrete placement as well as the strength of the steel section alone prior to composite action. Available Shear Strength According to AISC Specification Section I4.1, there are three acceptable methods for determining the available shear strength of the member: available shear strength of the steel section alone in accordance with Chapter G; available shear strength of the reinforced concrete portion alone per ACI 318; or available shear strength of the steel section in addition to the reinforcing steel ignoring the contribution of the concrete. Considering that the member in question does not have longitudinal reinforcing, it is determined by inspection that the shear strength will be controlled by the steel section alone using the provisions of Chapter G. From AISC Specification Section G4, the nominal shear strength, Vn, of box members is determined using AISC Specification Equation G4-1 with Cv2 determined from AISC Specification Section G2.2 with kv  5. As opposed to HSS sections that require the use of a reduced web area to take into account the corner radii, the web area of a box section may be used as follows: Aw  2 ht w , where h  clear distance between flanges  2  29.5 in.4 in.  14.8 in.2

The slenderness value, h/tw = h/t, which is the same as that calculated previously for use in local buckling classification,  = 118.

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 29, 000 ksi  1.37 kv E Fy  1.37 5    36 ksi   86.9  h t  118 Therefore, use AISC Specification Equation G2-11 to calculate Cv2. The web shear coefficient and nominal shear strength are calculated as:

Cv 2 

1.51kv E

(Spec. Eq. G2-11)

 h / tw 2 Fy 1.51 5 29,000 ksi   1182  36 ksi 

 0.437 Vn  0.6 Fy AwCv 2





 0.6  36 ksi  14.8 in.2  0.437 

(Spec. Eq. G4-1)

 140 kips

The available shear strength is checked as follows: LRFD

ASD

v  0.90

 v  1.67

vVn  0.90 140 kips 

Vn 140 kips  v 1.67  83.8 kips  22.1 kips

 126 kips  36.8 kips

o.k.

o.k.

Force Allocation and Load Transfer Load transfer calculations for applied axial forces should be performed in accordance with AISC Specification Section I6. The specific application of the load transfer provisions is dependent upon the configuration and detailing of the connecting elements. Expanded treatment of the application of load transfer provisions is provided in Example I.3. Summary

It has been determined that a 30 in. ~ 30 in. composite box column composed of 4-in.-thick plate is adequate for the imposed loads.

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EXAMPLE I.8 ENCASED COMPOSITE MEMBER FORCE ALLOCATION AND LOAD TRANSFER Given:

Refer to Figure I.8-1. Part I: For each loading condition (a) through (c), determine the required longitudinal shear force, Vr , to be transferred between the embedded steel section and concrete encasement. Part II: For loading condition (b), investigate the force transfer mechanisms of direct bearing and shear connection.

The composite member consists of an ASTM A992 W-shape encased by normal weight (145 lb/ft3) reinforced concrete having a specified concrete compressive strength, f c = 5 ksi. Deformed reinforcing bars conform to ASTM A615 with a minimum yield stress, Fyr, of 60 ksi. Applied loading, Pr, for each condition illustrated in Figure I.8-1 is composed of the following loads: PD = 260 kips PL = 780 kips

(a) External force to steel only

(b) External force to concrete only

(c) External force to both materials concurrently

Fig. I.8-1. Encased composite member in compression.

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Solution: Part I—Force Allocation

From AISC Manual Table 2-4, the steel material properties are: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1 and Figure I.8-1, the geometric properties of the encased W1045 are as follows: As  13.3 in.2 b f  8.02 in. t f  0.620 in. tw  0.350 in. d  10.1 in. h1  24 in. h2  24 in. Additional geometric properties of the composite section used for force allocation and load transfer are calculated as follows: Ag  h1h2   24 in. 24 in.  576 in.2 Asri  0.79 in.2 for a No. 8 bar n

Asr   Asri i 1



 8 0.79 in.2



 6.32 in.2

Ac  Ag  As  Asr  576 in.2  13.3 in.2  6.32 in.2  556 in.2 where Ac = cross-sectional area of concrete encasement, in.2 Ag = gross cross-sectional area of composite section, in.2 Asri = cross-sectional area of reinforcing bar i, in.2 Asr = cross-sectional area of continuous reinforcing bars, in.2 n = number of continuous reinforcing bars in composite section From ASCE/SEI 7, Chapter 2, the required strength is:

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LRFD

ASD

Pr  Pa  260 kips  780 kips  1, 040 kips

Pr  Pu  1.2  260 kips   1.6  780 kips   1,560 kips Composite Section Strength for Force Allocation

In accordance with AISC Specification Section I6, force allocation calculations are based on the nominal axial compressive strength of the encased composite member without length effects, Pno. This section strength is defined in Section I2.1b as: Pno  Fy As  Fysr Asr  0.85 f cAc



  50 ksi  13.3 in.

2

(Spec. Eq. I2-4)

   60 ksi   6.32 in.   0.85 5 ksi  556 in.  2

2

 3, 410 kips

Transfer Force for Condition (a) Refer to Figure I.8-1(a). For this condition, the entire external force is applied to the steel section only, and the provisions of AISC Specification Section I6.2a apply.  Fy As  Vr  Pr 1   Pno  

(Spec. Eq. I6-1)



  50 ksi  13.3 in.2  Pr 1   3, 410 kips   0.805 Pr

   

LRFD

Vr  0.805 1,560 kips 

ASD

Vr  0.805 1, 040 kips 

 1, 260 kips

 837 kips

Transfer Force for Condition (b) Refer to Figure I.8-1(b). For this condition, the entire external force is applied to the concrete encasement only, and the provisions of AISC Specification Section I6.2b apply.  Fy As  Vr  Pr    Pno    50 ksi  13.3 in.2  Pr   3, 410 kips   0.195 Pr



(Spec. Eq. I6-2a)

   

LRFD

Vr  0.195 1,560 kips   304 kips

ASD

Vr  0.195 1, 040 kips   203 kips

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Transfer Force for Condition (c) Refer to Figure I.8-1(c). For this condition, external force is applied to the steel section and concrete encasement concurrently, and the provisions of AISC Specification Section I6.2c apply. AISC Specification Commentary Section I6.2 states that when loads are applied to both the steel section and concrete encasement concurrently, Vr can be taken as the difference in magnitudes between the portion of the external force applied directly to the steel section and that required by Equation I6-2a. This concept can be written in equation form as follows:

 Fy As  Vr  Prs  Pr    Pno 

(Eq. 1)

where Prs = portion of external force applied directly to the steel section, kips Currently, the Specification provides no specific requirements for determining the distribution of the applied force for the determination of Prs, so it is left to engineering judgment. For a bearing plate condition such as the one represented in Figure I.8-1(c), one possible method for determining the distribution of applied forces is to use an elastic distribution based on the material axial stiffness ratios as follows: Ec  wc1.5 f c



 145 lb/ft 3



1.5

5 ksi

 3,900 ksi Es As   Prs    Pr  Es As  Ec Ac  Esr Asr 

 



  29, 000 ksi  13.3 in.2     29, 000 ksi  13.3 in.2   3,900 ksi  556 in.2   29, 000 ksi  6.32 in.2   0.141Pr











  Pr  

Substituting the results into Equation 1 yields:  Fy As  Vr  0.141Pr  Pr    Pno 



  50 ksi  13.3 in.2  0.141Pr  Pr   3, 410 kips   0.0540 Pr

   

LRFD

Vr  0.0540 1,560 kips   84.2 kips

ASD

Vr  0.0540 1, 040 kips   56.2 kips

An alternate approach would be use of a plastic distribution method whereby the load is partitioned to each material in accordance with their contribution to the composite section strength given in Equation I2-4. This method

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eliminates the need for longitudinal shear transfer provided the local bearing strength of the concrete and steel are adequate to resist the forces resulting from this distribution. Additional Discussion



The design and detailing of the connections required to deliver external forces to the composite member should be performed according to the applicable sections of AISC Specification Chapters J and K.



The connection cases illustrated by Figure I.8-1 are idealized conditions representative of the mechanics of actual connections. For instance, an extended single plate connection welded to the flange of the W10 and extending out beyond the face of concrete to attach to a steel beam is an example of a condition where it may be assumed that all external force is applied directly to the steel section only.

Solution: Part II—Load Transfer

The required longitudinal force to be transferred, Vr , determined in Part I condition (b) is used to investigate the applicable force transfer mechanisms of AISC Specification Section I6.3: direct bearing and shear connection. As indicated in the Specification, these force transfer mechanisms may not be superimposed; however, the mechanism providing the greatest nominal strength may be used. Note that direct bond interaction is not applicable to encased composite members as the variability of column sections and connection configurations makes confinement and bond strength more difficult to quantify than in filled HSS. Direct Bearing Determine Layout of Bearing Plates One method of utilizing direct bearing as a load transfer mechanism is through the use of internal bearing plates welded between the flanges of the encased W-shape as indicated in Figure I.8-2. When using bearing plates in this manner, it is essential that concrete mix proportions and installation techniques produce full bearing at the plates. Where multiple sets of bearing plates are used as illustrated in Figure I.8-2, it is recommended that the minimum spacing between plates be equal to the depth of the encased steel member to enhance constructability and concrete consolidation. For the configuration under consideration, this guideline is met with a plate spacing of 24 in.  d  10.1 in. Bearing plates should be located within the load introduction length given in AISC Specification Section I6.4a. The load introduction length is defined as two times the minimum transverse dimension of the composite member both above and below the load transfer region. The load transfer region is defined in Specification Commentary Section I6.4 as the depth of the connection. For the connection configuration under consideration, where the majority of the required force is being applied from the concrete column above, the depth of connection is conservatively taken as zero. Because the composite member only extends to one side of the point of force transfer, the bearing plates should be located within 2h2 = 48 in. of the top of the composite member as indicated in Figure I.8-2. Available Strength for the Limit State of Direct Bearing Assuming two sets of bearing plates are to be used as indicated in Figure I.8-2, the total contact area between the bearing plates and the concrete, A1, may be determined as follows:

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b f  tw 2 8.02 in.  0.350 in.  2  3.84 in.

a

b  d  2t f  10.1 in.  2  0.620 in.  8.86 in. c  width of clipped corners  w in.

Fig. I.8-2. Composite member with internal bearing plates.

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

A1  2ab  2c 2

  number of bearing plate sets 

2   2  3.84 in. 8.86 in.  2  w in.   2   

 134 in.2

The available strength for the direct bearing force transfer mechanism is: Rn  1.7 f cA1



 1.7  5 ksi  134 in.2



(Spec. Eq. I6-3)

 1,140 kips LRFD

ASD

B  0.65

B  2.31

B Rn  0.65 1,140 kips 

Rn 1,140 kips  B 2.31  494 kips  Vr  203 kips o.k.

 741 kips  Vr  304 kips o.k.

Thus, two sets of bearing plates are adequate. From these calculations, it can be seen that one set of bearing plates are adequate for force transfer purposes; however, the use of two sets of bearing plates serves to reduce the bearing plate thickness calculated in the following section. Required Bearing Plate Thickness There are several methods available for determining the bearing plate thickness. For rectangular plates supported on three sides, elastic solutions for plate stresses, such as those found in Roark’s Formulas for Stress and Strain (Young and Budynas, 2002), may be used in conjunction with AISC Specification Section F12 for thickness calculations. Alternately, yield line theory or computational methods such as finite element analysis may be employed. For this example, yield line theory is employed. Results of the yield line analysis depend on an assumption of column flange strength versus bearing plate strength in order to estimate the fixity of the bearing plate to column flange connection. In general, if the thickness of the bearing plate is less than the column flange thickness, fixity and plastic hinging can occur at this interface; otherwise, the use of a pinned condition is conservative. Ignoring the fillets of the W-shape and clipped corners of the bearing plate, the yield line pattern chosen for the fixed condition is depicted in Figure I.8-3. Note that the simplifying assumption of 45 yield lines illustrated in Figure I.8-3 has been shown to provide reasonably accurate results (Park and Gamble, 2000), and that this yield line pattern is only valid where b  2a. The plate thickness using Fy  36 ksi material may be determined as: LRFD

ASD

  0.90

  1.67

If t p  t f :

If t p  t f :

tp 

2a 2 wu  3b  2a  Fy  4a  b 

  t p    3Fy

  a 2 wa  3b  2a      4a  b     

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LRFD

ASD

If t p  t f :

If t p  t f : tp 

2a 2 wu  3b  2a 

  t p    3Fy

Fy  6a  b 

  a 2 wa  3b  2a         6a  b  

where wu  bearing pressure on plate determined

where wa  bearing pressure on plate determined

using LRFD load combinations Vr  A1

using ASD load combinations V  r A1



304 kips



134 in.2

203 kips 134 in.2

 2.27 ksi

 1.51 ksi

Assuming tp ≥ tf

Assuming tp ≥ tf

2  3.84 in.

tp 

 2.27 ksi   3  8.86 in.  2  3.84 in.     36 ksi   4  3.84 in.  8.86 in. 2

 0.733 in.

2 1.67  3.84 in. 1.51ksi  2

tp 

 3  8.86 in.  2  3.84 in.

3  36 ksi   4  3.84 in.  8.86 in.

 0.733 in.

Select w-in. plate. t p  w in.  t f  0.620 in. assumption o.k.

Select w-in. plate t p  w in.  t f  0.620 in. assumption o.k.

Thus, select w-in.-thick bearing plates.

Fig. I.8-3. Internal bearing plate yield line pattern (fixed condition).

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Bearing Plate to Encased Steel Member Weld The bearing plates should be connected to the encased steel member using welds designed in accordance with AISC Specification Chapter J to develop the full strength of the plate. For fillet welds, a weld size of stp will serve to develop the strength of either a 36- or 50-ksi plate as discussed in AISC Manual Part 10. Shear Connection Shear connection involves the use of steel headed stud or channel anchors placed on at least two faces of the steel shape in a generally symmetric configuration to transfer the required longitudinal shear force. For this example, win.-diameter ~ 4x-in.-long steel headed stud anchors composed of ASTM A108 material are selected. The specified minimum tensile strength, Fu, of ASTM A108 material is 65 ksi. Available Shear Strength of Steel Headed Stud Anchors The available shear strength of an individual steel headed stud anchor is determined in accordance with the composite component provisions of AISC Specification Section I8.3 as directed by Section I6.3b. Qnv  Fu Asa Asa 

  w in.

(Spec. Eq. I8-3) 2

4  0.442 in.2 LRFD

ASD

v  0.65

v  2.31



v Qnv  0.65  65 ksi  0.442 in.2





 18.7 kips per steel headed stud anchor

2 Qnv  65 ksi  0.442 in.  v 2.31



 12.4 kips per steel headed stud anchor

Required Number of Steel Headed Stud Anchors The number of steel headed stud anchors required to transfer the longitudinal shear is calculated as follows: LRFD

nanchors 

ASD

Vr v Qnv

nanchors 

304 kips 18.7 kips  16.3 steel headed stud anchors 

Vr Qnv v

203 kips 12.4 kips  16.4 steel headed stud anchors 

With anchors placed in pairs on each flange, select 20 anchors to satisfy the symmetry provisions of AISC Specification Section I6.4a. Placement of Steel Headed Stud Anchors Steel headed stud anchors are placed within the load introduction length in accordance with AISC Specification Section I6.4a. Because the composite member only extends to one side of the point of force transfer, the steel anchors are located within 2h2 = 48 in. of the top of the composite member.

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Placing two anchors on each flange provides four anchors per group, and maximum stud spacing within the load introduction length is determined as: smax 

load introduction length  distance to first anchor group from upper end of encased shape  total number of anchors   number of anchors per group   1  

48 in.  6 in.   20 anchors  4 anchors per group   1    10.5 in. 

Use 10 in. spacing beginning 6 in. from top of encased member. In addition to anchors placed within the load introduction length, anchors must also be placed along the remainder of the composite member at a maximum spacing of 32 times the anchor shank diameter = 24 in. in accordance with AISC Specification Sections I6.4a and I8.3e. The chosen anchor layout and spacing is illustrated in Figure I.8-4. Steel Headed Stud Anchor Detailing Limitations of AISC Specification Sections I6.4a, I8.1 and I8.3 Steel headed stud anchor detailing limitations are reviewed in this section with reference to the anchor configuration provided in Figure I.8-4 for anchors having a shank diameter, dsa, of w in. Note that these provisions are specific to the detailing of the anchors themselves and that additional limitations for the structural steel, concrete and reinforcing components of composite members should be reviewed as demonstrated in Design Example I.9. (1) Anchors must be placed on at least two faces of the steel shape in a generally symmetric configuration: Anchors are located in pairs on both faces. o.k. (2) Maximum anchor diameter: d sa  2.5  t f w in.  2.5  0.620 in.  1.55 in.



o.k.

(3) Minimum steel headed stud anchor height-to-diameter ratio: h / d sa  5 The minimum ratio of installed anchor height (base to top of head), h, to shank diameter, dsa, must meet the provisions of AISC Specification Section I8.3 as summarized in the User Note table at the end of the section. For shear in normal weight concrete the limiting ratio is five. As previously discussed, a 4x-in.-long anchor was selected from anchor manufacturer’s data. As the h/dsa ratio is based on the installed length, a length reduction for burn off during installation of x in. is taken to yield the final installed length of 4 in. h 4 in.   5.33  5 d sa w in.

o.k.

(4) Minimum lateral clear concrete cover = 12 in. From AWS D1.1 (AWS, 2015) Figure 7.1, the head diameter of a w-in.-diameter stud anchor is equal to 1.25 in.

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 h1   lateral spacing between anchor centerlines   anchor head diameter  lateral clear cover        2 2     2   24 in.   4 in.   1.25 in.       2   2   2   9.38 in.  12 in. o.k.

Fig. I.8-4. Composite member with steel anchors.

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(5) Minimum anchor spacing:

smin  4d sa  4  w in.  3.00 in. In accordance with AISC Specification Section I8.3e, this spacing limit applies in any direction. stransverse  4 in.  s min

o.k.

slongitudinal  10 in.  s min

o.k.

(6) Maximum anchor spacing: smax  32d sa  32  w in.  24.0 in.

In accordance with AISC Specification Section I6.4a, the spacing limits of Section I8.3e apply to steel anchor spacing both within and outside of the load introduction region. s  24.0 in.  smax

o.k.

(7) Clear cover above the top of the steel headed stud anchors: Minimum clear cover over the top of the steel headed stud anchors is not explicitly specified for steel anchors in composite components; however, in keeping with the intent of AISC Specification Section I1.1, it is recommended that the clear cover over the top of the anchor head follow the cover requirements of ACI 318 (ACI 318, 2014) Section 20.6.1. For concrete columns, ACI 318 specifies a clear cover of 12 in. h2

d  installed anchor length 2 2 24 in. 10.1 in.    4 in. 2 2  2.95 in.  12 in. o.k.

clear cover above anchor 



Concrete Breakout AISC Specification Section I8.3a states that in order to use Equation I8-3 for shear strength calculations as previously demonstrated, concrete breakout strength in shear must not be an applicable limit state. If concrete breakout is deemed to be an applicable limit state, the Specification provides two alternatives: either the concrete breakout strength can be determined explicitly using ACI 318, Chapter 17, in accordance with Specification Section I8.3a(b), or anchor reinforcement can be provided to resist the breakout force as discussed in Specification Section I8.3a(a). Determining whether concrete breakout is a viable failure mode is left to the engineer. According to AISC Specification Commentary Section I8.3, “it is important that it be deemed by the engineer that a concrete breakout failure mode in shear is directly avoided through having the edges perpendicular to the line of force supported, and the edges parallel to the line of force sufficiently distant that concrete breakout through a side edge is not deemed viable.”

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For the composite member being designed, no free edge exists in the direction of shear transfer along the length of the column, and concrete breakout in this direction is not an applicable limit state. However, it is still incumbent upon the engineer to review the possibility of concrete breakout through a side edge parallel to the line of force. One method for explicitly performing this check is through the use of the provisions of ACI 318, Chapter 17, as follows: ACI 318, Section 17.5.2.1(c), specifies that concrete breakout shall be checked for shear force parallel to the edge of a group of anchors using twice the value for the nominal breakout strength provided by ACI 318, Equation 17.5.2.1b, when the shear force in question acts perpendicular to the edge. For the composite member being designed, symmetrical concrete breakout planes form to each side of the encased shape, one of which is illustrated in Figure I.8-5.   0.75 for anchors governed by concrete breakout with supplemental reinforcement (provided by tie reinforcement) in accordance with ACI 318, Section 17.3.3

A  Vcbg  2  Vc  ec,V  ed ,V  c,V  h,V Vb  , for shear force parallel to an edge  AVco 

(ACI 318, Eq. 17.5.2.1b)

AVco  4.5  ca1 

(ACI 318, Eq. 17.5.2.1c)

2

 4.5 10 in.

2

 450 in.2

AVc  15 in.  40 in.  15 in. 24 in. , from Figure I.8-5  1, 680 in.2  ec,V  1.0 no eccentricity  ed ,V  1.0 in accordance with ACI 318, Section 17.5.2.1(c)  c ,V  1.4 compression-only member assumed uncracked  h ,V  1.0   l 0.2  Vb  8  e  da  a   d a  

f c  ca1 

1.5

(ACI 318, Eq. 17.5.2.3)

where le  4 in.  a-in. anchor head thickness from AWS D1.1, Figure 7.1

 3.63 in. d a  w-in. anchor diameter  a  1.0 from ACI 318, Section 17.2.6, for normal weight concrete   1.0 from ACI 318, Table 19.2.4.2, for normal weight concrete

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Vb

  3.63 in. 0.2  5, 000 psi = 8  w in.  1.0  10 in.1.5  1, 000 lb/kip   w in.    21.2 kips

1, 680 in.2  Vcbg  2  1.0 1.0 1.4 1.0  21.2 kips  2  450 in.   222 kips Vcbg  0.75  222 kips   167 kips per breakout plane Vcbg   2 breakout planes 167 kips/plane   334 kips Vcbg  Vr  304 kips o.k. Thus, concrete breakout along an edge parallel to the direction of the longitudinal shear transfer is not a controlling limit state, and Equation I8-3 is appropriate for determining available anchor strength.

Fig. I.8-5. Concrete breakout check for shear force parallel to an edge.

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Encased beam-column members with reinforcing detailed in accordance with the AISC Specification have demonstrated adequate confinement in tests to prevent concrete breakout along a parallel edge from occurring; however, it is still incumbent upon the engineer to review the project-specific detailing used for susceptibility to this limit state. If concrete breakout was determined to be a controlling limit state, transverse reinforcing ties could be analyzed as anchor reinforcement in accordance with AISC Specification Section I8.3a(a), and tie spacing through the load introduction length adjusted as required to prevent breakout. Alternately, the steel headed stud anchors could be relocated to the web of the encased member where breakout is prevented by confinement between the column flanges.

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EXAMPLE I.9 ENCASED COMPOSITE MEMBER IN AXIAL COMPRESSION Given: Determine if the encased composite member illustrated in Figure I.9-1 is adequate for the indicated dead and live loads.

Fig. I.9-1. Encased composite member section and applied loading. The composite member consists of an ASTM A992 W-shape encased by normal weight (145 lb/ft3) reinforced concrete having a specified concrete compressive strength, f c = 5 ksi. Deformed reinforcing bars conform to ASTM A615 with a minimum yield stress, Fyr, of 60 ksi. Solution: From AISC Manual Table 2-4, the steel material properties are: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, Figure I.9-1, and Design Example I.8, geometric and material properties of the composite section are: As h1 Ag Ec

= 13.3 in.2 = 24 in. = 576 in.2 = 3,900 ksi

bf = 8.02 in. h2 = 24 in. Asri = 0.790 in.2

tf = 0.620 in. Isx = 248 in.4 Asr = 6.32 in.2

d = 10.1 in. Isy = 53.4 in.4 Ac = 556 in.2

The moment of inertia of the reinforcing bars about the elastic neutral axis of the composite section, Isr, is required for composite member design and is calculated as follows:

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d b  1 in. for the diameter of a No. 8 bar

I sri  

db4 64  1 in.

4

64  0.0491 in.4 n

n

i 1

i 1

I sr   I sri   Asri ei 2



 







=8 0.0491 in.4  6 0.79 in.2  9.50 in.  2 0.79 in.2  0 in. 2

2

 428 in.4 where Asri = cross-sectional area of reinforcing bar i, in.2 Isri = moment of inertia of reinforcing bar i about its elastic neutral axis, in.4 Isr = moment of inertia of the reinforcing bars about the elastic neutral axis of the composite section, in.4 db = nominal diameter of reinforcing bar, in. ei = eccentricity of reinforcing bar i with respect to the elastic neutral axis of the composite section, in. n = number of reinforcing bars in composite section Note that the elastic neutral axis for each direction of the section in question is located at the x-x and y-y axes illustrated in Figure I.9-1, and that the moment of inertia calculated for the longitudinal reinforcement is valid about either axis due to symmetry. The moment of inertia values for the concrete about each axis are determined as:

I cx  I gx  I sx  I srx

 24 in.4

 248 in.4  428 in.4 12  27, 000 in.4 

I cy  I gy  I sy  I sry

 24 in.4

 53.4 in.4  428 in.4 12  27, 200 in.4 

Classify Section for Local Buckling In accordance with AISC Specification Section I1.2, local buckling effects need not be considered for encased composite members, thus all encased sections are treated as compact sections for strength calculations. Material and Detailing Limitations According to the User Note at the end of AISC Specification Section I1.1, the intent of the Specification is to implement the noncomposite detailing provisions of ACI 318 in conjunction with the composite-specific provisions of Specification Chapter I. Detailing provisions may be grouped into material related limits, transverse reinforcement provisions, and longitudinal and structural steel reinforcement provisions as illustrated in the following discussion.

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Material limits are provided in AISC Specification Sections I1.1(b) and I1.3 as follows: (1)

Concrete strength: f c  5 ksi o.k.

3 ksi  f c  10 ksi

(2) Specified minimum yield stress of structural steel:

Fy  75 ksi

Fy  50 ksi o.k. (3) Specified minimum yield stress of reinforcing bars:

Fyr  75 ksi

Fyr  60 ksi o.k. Transverse reinforcement limitations are provided in AISC Specification Section I1.1(c), I2.1a(b) and ACI 318 as follows: (1) Tie size and spacing limitations: The AISC Specification requires that either lateral ties or spirals be used for transverse reinforcement. Where lateral ties are used, a minimum of either No. 3 bars spaced at a maximum of 12 in. on center or No. 4 bars or larger spaced at a maximum of 16 in. on center are required. No. 3 lateral ties at 12 in. o.c. are provided. o.k. Note that AISC Specification Section I1.1(a) specifically excludes the composite column provisions of ACI 318, so it is unnecessary to meet the tie reinforcement provisions of ACI 318 when designing composite columns using the provisions of AISC Specification Chapter I. If spirals are used, the requirements of ACI 318 should be met according to the User Note at the end of AISC Specification Section I2.1a. (2) Additional tie size limitation: No. 4 ties or larger are required where No. 11 or larger bars are used as longitudinal reinforcement in accordance with ACI 318, Section 9.7.6.4.2. No. 3 lateral ties are provided for No. 8 longitudinal bars. o.k. (3) Maximum tie spacing should not exceed 0.5 times the least column dimension:  h1  24 in. smax  0.5 min   h2  24 in.  12.0 in. s  12.0 in.  smax

o.k.

(4) Concrete cover: ACI 318, Section 20.6.1.3 contains concrete cover requirements. For concrete not exposed to weather or in contact with ground, the required cover for column ties is 12 in.

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db  diameter of No. 3 tie 2  2.5 in.  2 in.  a in.

cover  2.5 in. 

 1.63 in.  12 in. o.k. (5) Provide ties as required for lateral support of longitudinal bars: AISC Specification Commentary Section I2.1a references ACI 318 for additional transverse tie requirements. In accordance with ACI 318, Section 25.7.2.3 and Figure R25.7.2.3a, ties are required to support longitudinal bars located farther than 6 in. clear on each side from a laterally supported bar. For corner bars, support is typically provided by the main perimeter ties. For intermediate bars, Figure I.9-1 illustrates one method for providing support through the use of a diamond-shaped tie. Longitudinal and structural steel reinforcement limits are provided in AISC Specification Sections I1.1, I2.1 and ACI 318 as follows: (1) Structural steel minimum reinforcement ratio:

As Ag  0.01

As 13.3 in.2   0.01 Ag 576 in.2  0.0231  0.01 o.k.

An explicit maximum reinforcement ratio for the encased steel shape is not provided in the AISC Specification; however, a range of 8 to 12% has been noted in the literature to result in economic composite members for the resistance of gravity loads (Leon and Hajjar, 2008). (2) Minimum longitudinal reinforcement ratio:

Asr Ag  0.004

Asr 6.32 in.2   0.004 Ag 576 in.2  0.0110  0.004 o.k. As discussed in AISC Specification Commentary Section I2.1a(c), only continuously developed longitudinal reinforcement is included in the minimum reinforcement ratio, so longitudinal restraining bars and other discontinuous longitudinal reinforcement is excluded. Note that this limitation is used in lieu of the minimum ratio provided in ACI 318 as discussed in Specification Commentary Section I1.1. (3) Maximum longitudinal reinforcement ratio:

Asr Ag  0.08

Asr 6.32 in.2   0.08 Ag 576 in.2  0.0110  0.08 o.k.

This longitudinal reinforcement limitation is provided in ACI 318, Section 10.6.1.1. It is recommended that all longitudinal reinforcement, including discontinuous reinforcement not used in strength calculations, be included in this ratio as it is considered a practical limitation to mitigate congestion of reinforcement. If longitudinal reinforcement is lap spliced as opposed to mechanically coupled, this limit is effectively reduced to 4% in areas away from the splice location. (4) Minimum number of longitudinal bars:

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ACI 318, Section 10.7.3.1, requires a minimum of four longitudinal bars within rectangular or circular members with ties and six bars for columns utilizing spiral ties. The intent for rectangular sections is to provide a minimum of one bar in each corner, so irregular geometries with multiple corners require additional longitudinal bars. 8 bars provided. o.k. (5) Clear spacing between longitudinal bars: ACI 318 Section 25.2.3 requires a clear distance between bars of 1.5db or 12 in. 1.5db  12 in. smin  max   12 in.    12 in. clear s  9.50 in.  1.00 in.  8.50 in.  12 in. o.k.

(6) Clear spacing between longitudinal bars and the steel core: AISC Specification Section I2.1e requires a minimum clear spacing between the steel core and longitudinal reinforcement of 1.5 reinforcing bar diameters, but not less than 12 in.

1.5db  12 in. smin  max    12 in.   12 in. clear Closest reinforcing bars to the encased section are the center bars adjacent to each flange: h2 d d   2.50 in.  b 2 2 2 24.0 in. 10.1 in. 1.00 in.    2.50 in.  2 2 2  3.95 in.  smin  12 in. o.k.

s

(7) Concrete cover for longitudinal reinforcement: ACI 318, Section 20.6.1.3, provides concrete cover requirements for reinforcement. The cover requirements for column ties and primary reinforcement are the same, and the tie cover was previously determined to be acceptable, thus the longitudinal reinforcement cover is acceptable by inspection. From ASCE/SEI, Chapter 2, the required compressive strength is: LRFD

Pr  Pu  1.2  260 kips   1.6  780 kips   1, 560 kips

ASD

Pr  Pa  260 kips  780 kips  1, 040 kips

Available Compressive Strength The nominal axial compressive strength without consideration of length effects, Pno, is determined from AISC Specification Section I2.1b as:

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Pno  Fy As  Fysr Asr  0.85 f cAc



  50 ksi  13.3 in.

2

(Spec. Eq. I2-4)

   60 ksi   6.32 in.   0.85 5 ksi  556 in.  2

2

 3, 410 kips

Because the unbraced length is the same in both the x-x and y-y directions, the column will buckle about the axis having the smaller effective composite section stiffness, EIeff. Noting the moment of inertia values determined previously for the concrete and reinforcing steel are similar about each axis, the column will buckle about the weak axis of the steel shape by inspection. Icy, Isy and Isry are therefore used for calculation of length effects in accordance with AISC Specification Section I2.1b as follows:  A  Asr C1  0.25  3  s  Ag

   0.7 

(Spec. Eq. I2-7)

 13.3 in.2  6.32 in.2   0.25  3    0.7 576 in.2    0.352  0.7; therefore C1  0.352 EI eff  Es I sy  Es I sry  C1 Ec I cy



(from Spec. Eq. I2-6)





  29, 000 ksi  53.4 in.4   29, 000 ksi  428 in.4



 0.352  3,900 ksi  27, 200 in.

4





2

 51,300, 000 kip-in.

Pe  2  EI eff  /  Lc  , where Lc  KL and K  1.0 for a pin-ended member 2





2 51,300, 000 kip-in.2

1.0 14 ft 12 in./ft    17,900 kips

(Spec. Eq. I2-5)



2

Pno 3, 410 kips  Pe 17,900 kips  0.191  2.25

Therefore, use AISC Specification Equation I2-2. Pno  Pn  Pno  0.658 Pe  

   

  3, 410 kips  0.658 

(Spec. Eq. I2-2) 0.191

 3,150 kips

Check adequacy of the composite column for the required axial compressive strength:

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LRFD

ASD

c  0.75

 c  2.00

c Pn  0.75  3,150 kips 

Pn 3,150 kips  2.00 c  1,580 kips  1,040 kips o.k.

 2,360 kips  1,560 kips

o.k.

Available Compressive Strength of Composite Section Versus Bare Steel Section Due to the differences in resistance and safety factors between composite and noncomposite column provisions, it is possible in rare instances to calculate a lower available compressive strength for an encased composite column than one would calculate for the corresponding bare steel section. However, in accordance with AISC Specification Section I2.1b, the available compressive strength need not be less than that calculated for the bare steel member in accordance with Chapter E. From AISC Manual Table 4-1a: LRFD

c Pn  359 kips  2, 360 kips

ASD

Pn  239 kips  1, 580 kips c

Thus, the composite section strength controls and is adequate for the required axial compressive strength as previously demonstrated. Force Allocation and Load Transfer Load transfer calculations for external axial forces should be performed in accordance with AISC Specification Section I6. The specific application of the load transfer provisions is dependent upon the configuration and detailing of the connecting elements. Expanded treatment of the application of load transfer provisions for encased composite members is provided in Design Example I.8. Typical Detailing Convention Designers are directed to AISC Design Guide 6 (Griffis, 1992) for additional discussion and typical details of encased composite columns not explicitly covered in this example.

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EXAMPLE I.10

ENCASED COMPOSITE MEMBER IN AXIAL TENSION

Given: Determine if the encased composite member illustrated in Figure I.10-1 is adequate for the indicated dead load compression and wind load tension. The entire load is applied to the encased steel section.

Fig. I.10-1. Encased composite member section and applied loading. The composite member consists of an ASTM A992 W-shape encased by normal weight (145 lb/ft3) reinforced concrete having a specified concrete compressive strength, f c = 5 ksi. Deformed reinforcing bars conform to ASTM A615 with a minimum yield stress, Fyr, of 60 ksi.

Solution: From AISC Manual Table 2-4, the steel material properties are: ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1 and Figure I.10-1, the relevant properties of the composite section are: As = 13.3 in.2 Asr = 6.32 in.2 (area of eight No. 8 bars) Material and Detailing Limitations Refer to Design Example I.9 for a check of material and detailing limitations specified in AISC Specification Chapter I for encased composite members.

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Taking compression as negative and tension as positive, from ASCE/SEI 7, Chapter 2, the required strength is: LRFD

ASD

Governing uplift load combination  0.9D  1.0W

Governing uplift load combination  0.6D  0.6W

Pr  Pu

Pr  Pa

 0.9  260 kips   1.0  980 kips 

 0.6  260 kips   0.6  980 kips 

 746 kips

 432 kips

Available Tensile Strength Available tensile strength for an encased composite member is determined in accordance with AISC Specification Section I2.1c. Pn  Fy As  Fysr Asr



(Spec. Eq. I2-8)





  50 ksi  13.3 in.2   60 ksi  6.32 in.2



 1, 040 kips LRFD

ASD

t  0.90

t  1.67

t Pn  0.90 1, 040 kips 

Pn 1, 040 kips  t 1.67

 936 kips  746 kips

o.k.

 623 kips  432 kips

o.k.

Force Allocation and Load Transfer In cases where all of the tension is applied to either the reinforcing steel or the encased steel shape, and the available strength of the reinforcing steel or encased steel shape by itself is adequate, no additional load transfer calculations are required. In cases, such as the one under consideration, where the available strength of both the reinforcing steel and the encased steel shape are needed to provide adequate tension resistance, AISC Specification Section I6 can be modified for tensile load transfer requirements by replacing the Pno term in Equations I6-1 and I6-2 with the nominal tensile strength, Pn, determined from Equation I2-8. For external tensile force applied to the encased steel section:  Fy As  Vr  Pr 1   Pn  

(Spec. Eq. C-I6-1)

For external tensile force applied to the longitudinal reinforcement of the concrete encasement:  Fy As  Vr  Pr    Pn 

(Spec. Eq. C-I6-2)

where Pn = nominal tensile strength of encased composite member from Equation I2-8, kips Pr = required external tensile force applied to the composite member, kips

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Per the problem statement, the entire external force is applied to the encased steel section, thus, AISC Specification Equation C-I6-1 is used as follows:



  50 ksi  13.3 in.2 Vr  Pr 1   1, 040 kips   0.361Pr

   

LRFD

Vr  0.361 746 kips   269 kips

ASD

Vr  0.361 432 kips   156 kips

The longitudinal shear force must be transferred between the encased steel shape and longitudinal reinforcing using the force transfer mechanisms of direct bearing or shear connection in accordance with AISC Specification Section I6.3 as illustrated in Example I.8.

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EXAMPLE I.11 ENCASED COMPOSITE MEMBER IN COMBINED AXIAL COMPRESSION, FLEXURE AND SHEAR Given: Determine if the encased composite member illustrated in Figure I.11-1 is adequate for the indicated axial forces, shears and moments that have been determined in accordance with the direct analysis method of AISC Specification Chapter C for the controlling ASCE/SEI 7 load combinations.

Fig. I.11-1. Encased composite member section and member forces. The composite member consists of an ASTM A992 W-shape encased by normal weight (145 lb/ft3) reinforced concrete having a specified concrete compressive strength, f c = 5 ksi. Deformed reinforcing bars conform to ASTM A615 with a minimum yield stress, Fyr, of 60 ksi.

Solution: From AISC Manual Table 2-4, the steel material properties are: ASTM A992 Fy = 50 ksi Fu = 65 ksi

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From AISC Manual Table 1-1, Figure I.11-1, and Examples I.8 and I.9, the geometric and material properties of the composite section are: As = 13.3 in.2 Ag = 576 in.2 Ac = 556 in.2 Asr = 6.32 in.2 c = 22 in.

d bf tf tw Ssx

= 10.1 in. = 8.02 in. = 0.620 in. = 0.350 in. = 49.1 in.3

h1 = 24 in. h2 = 24 in. Ec = 3,900 ksi Zsx = 54.9 in.3

Isy Icx Icy Isr

= 53.4 in.4 = 27,000 in.4 = 27,200 in.4 = 428 in.4

The area of continuous reinforcing located at the centerline of the composite section, Asrs, is determined from Figure I.11-1 as follows:

Asrs  2  Asrsi 



 2 0.79 in.2



 1.58 in.2 where Asrsi  area of reinforcing bar i at centerline of composite section

 0.79 in.2 for a No. 8 bar For the section under consideration, Asrs is equal about both the x-x and y-y axis. Classify Section for Local Buckling In accordance with AISC Specification Section I1.2, local buckling effects need not be considered for encased composite members, thus all encased sections are treated as compact sections for strength calculations. Material and Detailing Limitations Refer to Design Example I.9 for a check of material and detailing limitations. Interaction of Axial Force and Flexure Interaction between flexure and axial forces in composite members is governed by AISC Specification Section I5, which permits the use of the methods outlined in Section I1.2. The strain compatibility method is a generalized approach that allows for the construction of an interaction diagram based upon the same concepts used for reinforced concrete design. Application of the strain compatibility method is required for irregular/nonsymmetrical sections, and its general implementation may be found in reinforced concrete design texts and will not be discussed further here. Plastic stress distribution methods are discussed in AISC Specification Commentary Section I5, which provides four procedures applicable to encased composite members. The first procedure, Method 1, invokes the interaction equations of Section H1. The second procedure, Method 2, involves the construction of a piecewise-linear interaction curve using the plastic strength equations provided in AISC Manual Table 6-3a. The third procedure, Method 2—Simplified, is a reduction of the piecewise-linear interaction curve that allows for the use of less conservative interaction equations than those presented in Chapter H. The fourth and final procedure, Method 3, utilizes AISC Design Guide 6 (Griffis, 1992). For this design example, three of the available plastic stress distribution procedures are reviewed and compared. Method 3 is not demonstrated as it is not applicable to the section under consideration due to the area of the encased steel section being smaller than the minimum limit of 4% of the gross area of the composite section provided in the earlier Specification upon which Design Guide 6 is based.

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Method 1—Interaction Equations of Section H1 The most direct and conservative method of assessing interaction effects is through the use of the interaction equations of AISC Specification Section H1. Unlike concrete filled HSS shapes, the available compressive and flexural strengths of encased members are not tabulated in the AISC Manual due to the large variety of possible combinations. Calculations must therefore be performed explicitly using the provisions of Chapter I. Available Compressive Strength The available compressive strength is calculated as illustrated in Example I.9. LRFD

ASD

c Pn  2, 360 kips

Pn  1, 580 kips c

Nominal Flexural Strength The applied moment illustrated in Figure I.11-1 is resisted by the flexural strength of the composite section about its strong (x-x) axis. The strength of the section in pure flexure is calculated using the equations of AISC Manual Table 6-3a for Point B. Note that the calculation of the flexural strength at Point B first requires calculation of the flexural strength at Point D as follows: h  Z r   Asr  Asrs   2  c  2  





 24 in.   6.32 in.2  1.58 in.2   22 in.   2   45.0 in.3

Zc  

h1h 22 4

 Zs  Zr

 24 in. 24 in.2

4  3,360 in.3

 54.9 in.3  45.0 in.3

Z  M D  Fy Z s  Fyr Z r  0.85 f c  c   2    3,360 in.3    1    50 ksi  54.9 in.3   60 ksi  45.0 in.3  0.85  5 ksi       2 12 in./ft      









 1, 050 kip-ft d d Assuming hn is within the flange   t f  hn   : 2 2 

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hn 

0.85 f c  Ac  As  db f  Asrs   2 Fy  As  db f   2 Fyr Asrs 2 0.85 f c  h1  b f   2 Fy b f 

0.85  5 ksi  556 in.2  13.3 in.2  10.1 in. 8.02 in.  1.58 in.2      2  2  50 ksi  13.3 in.  10.1 in. 8.02 in.   2  60 ksi  1.58 in.2    2  0.85  5 ksi   24 in.  8.02 in.  2  50 ksi  8.02 in. 





    

 4.98 in.

Check assumption: 10.1 in.  10.1 in.   0.620 in.   hn   2  2  4.43 in.  hn  4.98 in.  5.05 in. assumption o.k. d  d  Z sn  Z s  b f   hn   hn  2  2   10.1 in.   10.1 in.   54.9 in.3   8.02 in.   4.98 in.    4.98 in.  2 2     49.3 in.3

Z cn  h1h 2n  Z sn   24 in. 4.98 in.  49.3 in.3 2

 546 in.3 Z  M B  M D  Fy Z sn  0.85 fc  cn   2    546 in.3    1   12, 600 kip-in.   50 ksi  49.3 in.3  0.85  5 ksi       2      12 in./ft   748 kip-ft





Available Flexural Strength LRFD

ASD

b  0.90

 b  1.67

b M n  0.90  748 kip-ft 

M n 748 kip-ft  b 1.67  448 kip-ft

 673 kip-ft

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Interaction of Axial Compression and Flexure LRFD

ASD

Pn /  c  1, 580 kips M n /  c  448 kip-ft

c Pn  2,360 kips b M n  673 kip-ft

Pr Pa  Pc Pn /  c 879 kips  1, 580 kips

Pr P  u Pc c Pn 1,170 kips 2,360 kips  0.496  0.2 

 0.556  0.2

Therefore, use AISC Specification Equation H1-1a. Pu 8  Mu      1.0 c Pn 9  b M n  8  670 kip-ft  0.496     1.0 9  673 kip-ft  1.38  1.0

n.g.

(from Spec. Eq. H1-1a)

Therefore, use AISC Specification Equation H1-1a. Pa 8  Ma      1.0 Pn / c 9  M n / b 

(from Spec. Eq. H1-1a)

8  302 kip-ft  0.556     1.0 9  448 kip-ft  1.16  1.0 n.g.

Method 1 indicates that the section is inadequate for the applied loads. The designer can elect to choose a new section that passes the interaction check or re-analyze the current section using a less conservative design method such as Method 2. The use of Method 2 is illustrated in the following section. Method 2—Interaction Curves from the Plastic Stress Distribution Model The procedure for creating an interaction curve using the plastic stress distribution model is illustrated graphically in AISC Specification Commentary Figure C-I5.2, and repeated here.

Fig. C-I5.2. Interaction diagram for composite beam-columns—Method 2.

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Referencing Figure C.I5.2, the nominal strength interaction surface A, B, C, D is first determined using the equations of AISC Manual Table 6-3a. This curve is representative of the short column member strength without consideration of length effects. A slenderness reduction factor, , is then calculated and applied to each point to create surface A , B, C, D . The appropriate resistance or safety factors are then applied to create the design surface A , B , C , D . Finally, the required axial and flexural strengths from the applicable load combinations of ASCE/SEI 7 are plotted on the design surface. The member is then deemed acceptable for the applied loading if all points fall within the design surface. These steps are illustrated in detail by the following calculations. Step 1: Construct nominal strength interaction surface A, B, C, D without length effects Using the equations provided in Figure I-1a for bending about the x-x axis yields: Point A (pure axial compression): PA  Fy As  Fyr Asr  0.85 f cAc











  50 ksi  13.3 in.2   60 ksi  6.32 in.2  0.85  5 ksi  556 in.2



 3, 410 kips

M A  0 kip-ft Point D (maximum nominal moment strength):

PD  

0.85 f cAc 2



0.85  5 ksi  556 in.2



2

 1,180 kips Calculation of MD was demonstrated previously in Method 1. M D  1, 050 kip-ft

Point B (pure flexure): PB  0 kips

Calculation of MB was demonstrated previously in Method 1. M B  748 kip-ft

Point C (intermediate point): PC  0.85 f cAc



 0.85  5 ksi  556 in.2



 2,360 kips MC  M B  748 kip-ft

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The calculated points are plotted to construct the nominal strength interaction surface without length effects as depicted in Figure I.11-2. Step 2: Construct nominal strength interaction surface A , B, C, D with length effects The slenderness reduction factor, , is calculated for Point A using AISC Specification Section I2.1 in accordance with AISC Specification Commentary Section I5. Because the unbraced length is the same in both the x-x and y-y directions, the column will buckle about the axis having the smaller effective composite section stiffness, EIeff. Noting the moment of inertia values for the concrete and reinforcing steel are similar about each axis, the column will buckle about the weak axis of the steel shape by inspection. Icy, Isy and Isry are therefore used for calculation of length effects in accordance with AISC Specification Section I2.1b. Pno  PA  3, 410 kips  As  Asr C1  0.25  3   Ag

   0.7 

(Spec. Eq. I2-7)

 13.3 in.2  6.32 in.2   0.25  3    0.7 576 in.2    0.352  0.7; therefore C1  0.352. EI eff  Es I sy  Es I sry  C1 Ec I cy



  29, 000 ksi  53.4 in.

4

(from Spec. Eq. I2-6)

   29, 000 ksi   428 in.   0.352 3,900 ksi   27, 200 in.  4

4

 51,300, 000 kip-in.2

Fig. I.11-2. Nominal strength interaction surface without length effects.

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Pe  2  EI eff  /  Lc  , where Lc  KL and K  1.0 2

(Spec. Eq. I2-5)

in accordance with the direct analysis method 



2 51, 300, 000 kip-in.2

1.0 14 ft 12 in./ft    17,900 kips



2

Pno 3, 410 kips  Pe 17,900 kips  0.191  2.25

Therefore, use AISC Specification Equation I2-2. Pno   Pn  Pno 0.658 Pe  

   

  3, 410 kips  0.658 

(Spec. Eq. I2-2) 0.191

 3,150 kips Pn Pno 3,150 kips  3, 410 kips



 0.924

In accordance with AISC Specification Commentary Section I5, the same slenderness reduction is applied to each of the remaining points on the interaction surface as follows: PA  PA  0.924  3, 410 kips   3,150 kips PB  PB  0.924  0 kip   0 kip PC   PC  0.924  2,360 kips   2,180 kips PD  PD  0.924 1,180 kips   1, 090 kips

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The modified axial strength values are plotted with the flexural strength values previously calculated to construct the nominal strength interaction surface including length effects. These values are superimposed on the nominal strength surface not including length effects for comparison purposes in Figure I.11-3. The consideration of length effects results in a vertical reduction of the nominal strength curve as illustrated by Figure I.11-3. This vertical movement creates an unsafe zone within the shaded area of the figure where flexural capacities of the nominal strength (with length effects) curve exceed the section capacity. Application of resistance or safety factors reduces this unsafe zone as illustrated in the following step; however, designers should be cognizant of the potential for unsafe designs with loads approaching the predicted flexural capacity of the section. Alternately, the use of Method 2—Simplified eliminates this possibility altogether. Step 3: Construct design interaction surface A, B, C, D and verify member adequacy The final step in the Method 2 procedure is to reduce the interaction surface for design using the appropriate resistance or safety factors. The available compressive and flexural strengths are determined as follows: LRFD

ASD

c  0.75

 c  2.00

PX   c PX  , where X  A, B, C or D

PX  

PA  0.75  3,150 kips 

PA  3,150 kips / 2.00

 2,360 kips PB  0.75  0 kip   0 kip PC   0.75  2,180 kips   1, 640 kips PD  0.75 1, 090 kips   818 kips

PX  , where X  A, B, C or D c

 1,580 kips PB  0 kip / 2.00  0 kip PC   2,180 kips / 2.00  1, 090 kips PD  1, 090 kips / 2.00  545 kips

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LRFD

ASD

b  0.90

 b  1.67

M X   b M X , where X  A, B, C or D

M X  

M A  0.90  0 kip-ft 

M A  0 kip-ft /1.67

 0 kip-ft

MX , where X  A, B, C or D b

 0 kip-ft

M B  0.90  748 kip-ft 

M B  748 kip-ft /1.67

 673 kip-ft

 448 kip-ft

M C   0.90  748 kip-ft 

M C   748 kip-ft /1.67

 673 kip-ft

 448 kip-ft

M D  0.90 1, 050 kip-ft   945 kip-ft

M D  1, 050 kip-ft /1.67  629 kip-ft

The available strength values for each design method can now be plotted. These values are superimposed on the nominal strength surfaces (with and without length effects) previously calculated for comparison purposes in Figure I.11-4. By plotting the required axial and flexural strength values on the available strength surfaces indicated in Figure I.11-4, it can be seen that both ASD (Ma,Pa) and LRFD (Mu,Pu) points lie within their respective design surfaces. The member in question is therefore adequate for the applied loads.

Fig. I.11-3. Nominal strength interaction surfaces (with and without length effects).

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As discussed previously in Step 2 as well as in AISC Specification Commentary Section I5, when reducing the flexural strength of Point D for length effects and resistance or safety factors, an unsafe situation could result whereby additional flexural strength is permitted at a lower axial compressive strength than predicted by the cross section strength of the member. This effect is highlighted by the magnified portion of Figure I.11-4, where LRFD design point D closely approaches the nominal strength curve. Designs falling outside the nominal strength curve are unsafe and not permitted. Method 2—Simplified The unsafe zone discussed in the previous section for Method 2 is avoided in the Method 2—Simplified procedure by the removal of Point D from the Method 2 interaction surface leaving only points A, B and C as illustrated in Figure I.11-5. Reducing the number of interaction points also allows for a bilinear interaction check defined by AISC Specification Commentary Equations C-I5-1a and C-I5-1b to be performed. Using the available strength values previously calculated in conjunction with the Commentary equations, interaction ratios are determined as follows: LRFD

ASD

Pr  Pu  1,170 kips  PC   1, 640 kips

Pr  Pa  879 kips  PC   1, 090 kips

Therefore, use AISC Specification Commentary Equation C-I5-1a.

Therefore, use AISC Specification Commentary Equation C-I5-1a.

Mr Mu   1.0 M C M C  670 kip-ft  1.0 673 kip-ft

Mr Ma   1.0 M C M C 

1.0  1.0

(from Spec. Comm. Eq. C-I5-1a)

(from Spec. Comm. Eq. C-I5-1a)

302 kip-ft  1.0 448 kip-ft 0.67  1.0 o.k.

o.k.

Thus, the member is adequate for the applied loads. Comparison of Methods The composite member was found to be inadequate using Method 1—Chapter H interaction equations, but was found to be adequate using both Method 2 and Method 2—Simplified procedures. A comparison between the methods is most easily made by overlaying the design curves from each method as illustrated in Figure I.11-6 for LRFD design. From Figure I.11-6, the conservative nature of the Chapter H interaction equations can be seen. Method 2 provides the highest available strength; however, the Method 2—Simplified procedure also provides a good representation of the design curve. The procedure in Figure I-1 for calculating the flexural strength of Point C first requires the calculation of the flexural strength for Point D. The design effort required for the Method 2—Simplified procedure, which utilizes Point C, is therefore not greatly reduced from Method 2. Available Shear Strength According to AISC Specification Section I4.1, there are three acceptable options for determining the available shear strength of an encased composite member: (1) Option 1—Available shear strength of the steel section alone in accordance with AISC Specification Chapter G. (2) Option 2—Available shear strength of the reinforced concrete portion alone per ACI 318.

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(3) Option 3—Available shear strength of the steel section, in addition to the reinforcing steel ignoring the contribution of the concrete.

Fig. I.11-4. Available and nominal interaction surfaces.

Fig. I.11-5. Comparison of Method 2 and Method 2 —Simplified.

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Option 1—Available Shear Strength of Steel Section A W1045 member meets the criteria of AISC Specification Section G2.1(a) according to the User Note at the end of the section. As demonstrated in Design Example I.9, No. 3 ties at 12 in. on center as illustrated in Figure I.11-1 satisfy the minimum detailing requirements of the Specification. The nominal shear strength may therefore be determined as: Cv1  1.0

(Spec. Eq. G2-2)

Aw  dtw  10.1 in. 0.350 in.  3.54 in.2 Vn  0.6 Fy AwCv1

(Spec. Eq. G2-1)





 0.6  50 ksi  3.54 in.2 1.0   106 kips

The available shear strength of the steel section is: LRFD

ASD

v  1.00

 v  1.50

vVn  1.00 106 kips 

Vn 106 kips  v 1.50  70.7 kips  57.4 kips

 106 kips  95.7 kips

o.k.

Fig. I.11-6. Comparison of interaction methods (LRFD).

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Option 2—Available Shear Strength of the Reinforced Concrete (Concrete and Transverse Steel Reinforcement) The available shear strength of the steel section alone has been shown to be sufficient; however, the amount of transverse reinforcement required for shear resistance in accordance with AISC Specification Section I4.1(b) will be determined for demonstration purposes. Tie Requirements for Shear Resistance The nominal concrete shear strength is: Vc  2 f cbw d

(ACI 318, Eq. 22.5.5.1)

where   1.0 for normal weight concrete from ACI 318, Table 19.2.4.2

bw  h1 d  distance from extreme compression fiber to centroid of longitudinal tension reinforcement  24 in.  22 in.  21.5 in.  1 kip  Vc  2 1.0  5, 000 psi  24 in. 21.5 in.    1, 000 lb   73.0 kips The tie requirements for shear resistance are determined from ACI 318 Chapter 22 and AISC Specification Section I4.1(b), as follows: LRFD

ASD

 v  2.00

 v  0.75

Av Vu  vVc  s v f yr d



(from ACI 318, Eq. R22.5.10.5)

95.7 kips  0.75  73.0 kips 

 0.0423 in.

Using two legs of No. 3 ties with Av = 0.11 in.2 from ACI 318, Appendix A:



s s  5.20 in.

  0.0423 in.



s s  9.46 in.

  0.0423 in.

Using two legs of No. 3 ties with Av = 0.11 in.2 from ACI 318, Appendix A:



2 0.11 in.2 s s  6.79 in.

Using two legs of the No. 4 ties with Av = 0.20 in.2: 2 0.20 in.2

(from ACI 318, Eq. R22.5.10.5)

 73.0 kips  57.4 kips     2.00    60 ksi  21.5 in. 2.00  0.0324 in.

0.75  60 ksi  21.5 in.

2 0.11 in.2

Av Va  Vc v   s f yr d v

  0.0324 in.

Using two legs of the No. 4 ties with Av = 0.20 in.2:



2 0.20 in.2 s s  12.3 in.

  0.0324 in.

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LRFD

ASD

From ACI 318, Section 9.7.6.2.2, the maximum spacing From ACI 318, Section 9.7.6.2.2, the maximum spacing is: is: d d smax  smax  2 2 21.5 in. 21.5 in.   2 2  10.8 in.  10.8 in. Use No. 3 ties at 5 in. o.c. or No. 4 ties at 9 in. o.c.

Use No. 3 ties at 6 in. o.c. or No. 4 ties at 10 in. o.c.

Minimum Reinforcing Limits Check that the minimum shear reinforcement is provided as required by ACI 318, Section 9.6.3.3. Av ,min s

b  0.75 f c  w  f yr 

 50bw   f yr 

0.75 5, 000 psi  24 in. 60, 000 psi

(ACI 318, Table 9.6.3.3) 

50  24 in. 60, 000 psi

 0.0212 in.  0.0200 in. LRFD

ASD

Av  0.0423 in.  0.0212 in. o.k. s

Av  0.0324 in.  0.0212 in. o.k. s

Maximum Reinforcing Limits From ACI 318, Section 9.7.6.2.2, maximum stirrup spacing is reduced to d/4 if Vs  4 f cbw d . If No. 4 ties at 9 in. on center are selected: Vs  

Av f yr d



s

2 0.20 in.2

(ACI 318, Eq. 22.5.10.5.3)

  60 ksi  21.5 in. 9 in.

 57.3 kips Vs ,max  4 f cbw d  1 kip   4 5, 000 psi  24 in. 21.5 in.    1, 000 lb   146 kips  57.3 kips

Therefore, the stirrup spacing is acceptable. Option 3—Determine Available Shear Strength of the Steel Section plus Reinforcing Steel The third procedure combines the shear strength of the reinforcing steel with that of the encased steel section, ignoring the contribution of the concrete. AISC Specification Section I4.1(c) provides a combined resistance and safety factor for this procedure. Note that the combined resistance and safety factor takes precedence over the

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factors in Chapter G used for the encased steel section alone in Option 1. The amount of transverse reinforcement required for shear resistance is determined as follows: Tie Requirements for Shear Resistance The nominal shear strength of the encased steel section was previously determined to be: Vn , steel  106 kips

The tie requirements for shear resistance are determined from ACI 318, Chapter 22, and AISC Specification Section I4.1(c), as follows: LRFD

ASD

v  0.75 Av s





v  2.00 Av Va  Vn, steel v   s f yr d v

Vu  vVn, steel v f yr d

95.7 kips  0.75 106 kips 



0.75  60 ksi  21.5 in.

57.4 kips  106 kips 2.00 

  60 ksi  21.5 in.    2.00    0.00682 in.

 0.0167 in.

As determined in Option 2, the minimum value of Av s  0.0212 , and the maximum tie spacing for shear resistance is 10.8 in. Using two legs of No. 3 ties for Av:



2 0.11 in.2

  0.0212 in.

s s  10.4 in.  smax  10.8 in. Use No. 3 ties at 10 in. o.c. Summary and Comparison of Available Shear Strength Calculations The use of the steel section alone is the most expedient method for calculating available shear strength and allows the use of a tie spacing which may be greater than that required for shear resistance by ACI 318. Where the strength of the steel section alone is not adequate, Option 3 will generally result in reduced tie reinforcement requirements as compared to Option 2. Force Allocation and Load Transfer Load transfer calculations should be performed in accordance with AISC Specification Section I6. The specific application of the load transfer provisions is dependent upon the configuration and detailing of the connecting elements. Expanded treatment of the application of load transfer provisions for encased composite members is provided in Design Example I.8 and AISC Design Guide 6.

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EXAMPLE I.12 STEEL ANCHORS IN COMPOSITE COMPONENTS Given: Select an appropriate w-in.-diameter, Type B steel headed stud anchor to resist the dead and live loads indicated in Figure I.12-1. The anchor is part of a composite system that may be designed using the steel anchor in composite components provisions of AISC Specification Section I8.3.

Fig. I.12-1. Steel headed stud anchor and applied loading. The steel headed stud anchor is encased by normal weight (145 lb/ft3) reinforced concrete having a specified concrete compressive strength, f c = 5 ksi. In accordance with AISC Manual Part 2, headed stud anchors shall be in accordance with AWS D1.1 with a specified minimum tensile stress, Fu, of 65 ksi. The anchor is located away from edges such that concrete breakout in shear is not a viable limit state, and the nearest anchor is located 24 in. away. The concrete is considered to be uncracked.

Solution: Minimum Anchor Length AISC Specification Section I8.3 provides minimum length to shank diameter ratios for anchors subjected to shear, tension, and interaction of shear and tension in both normal weight and lightweight concrete. These ratios are also summarized in the User Note provided within Section I8.3. For normal weight concrete subject to shear and tension, h / d sa  8 , thus:

h  8d sa  8  w in.  6.00 in. This length is measured from the base of the steel headed stud anchor to the top of the head after installation. From anchor manufacturer’s data, a standard stock length of 6x in. is selected. Using a x-in. length reduction to account for burn off during installation yields a final installed length of 6.00 in. 6.00 in.  6.00 in.

o.k.

Select a w-in.-diameter  6x-in.-long headed stud anchor.

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Required Shear and Tensile Strength From ASCE/SEI 7, Chapter 2, the required shear and tensile strengths are: LRFD

ASD

Governing load combination for interaction = 1.2D + 1.6L

Governing load combination for interaction =D+L

Quv  1.2  2 kips   1.6  5 kips 

Qav  2 kips  5 kips  7.00 kips (shear)

 10.4 kips (shear)

Qat  3 kips  7.5 kips  10.5 kips (tension)

Qut  1.2  3 kips   1.6  7.5 kips   15.6 kips (tension) Available Shear Strength

Per the problem statement, concrete breakout is not considered to be an applicable limit state. AISC Equation I8-3 may therefore be used to determine the available shear strength of the steel headed stud anchor as follows: Qnv  Fu Asa

(Spec. Eq. I8-3)

where Asa  cross-sectional area of steel headed stud anchor



  w in.

2

4  0.442 in.2



Qnv   65 ksi  0.442 in.2



 28.7 kips LRFD

ASD

v  0.65

 v  2.31

v Qnv  0.65  28.7 kips 

Qnv 28.7 kips  v 2.31

 18.7 kips

 12.4 kips

Alternately, available shear strengths can be selected directly from Table I.12-1 located at the end of this example. Available Tensile Strength The nominal tensile strength of a steel headed stud anchor is determined using AISC Specification Equation I8-4 provided the edge and spacing limitations of AISC Specification Section I8.3b are met as follows: (1) Minimum distance from centerline of anchor to free edge: 1.5h  1.5  6.00 in.  9.00 in. There are no free edges, therefore this limitation does not apply.

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(2) Minimum distance between centerlines of adjacent anchors: 3h  3  6.00 in.  18.0 in. 18.0 in.  24 in.

o.k.

Equation I8-4 may therefore be used as follows: Qnt  Fu Asa



  65 ksi  0.442 in.

2

(Spec. Eq. I8-4)



 28.7 kips LRFD

ASD

t  0.75

 t  2.00

t Qnt  0.75  28.7 kips 

Qnt 28.7 kips  t 2.00

 21.5 kips

 14.4 kips

Alternately, available tensile strengths can be selected directly from Table I.12-1 located at the end of this example. Interaction of Shear and Tension The detailing limits on edge distances and spacing imposed by AISC Specification Section I8.3c for shear and tension interaction are the same as those previously reviewed separately for tension and shear alone. Tension and shear interaction is checked using Specification Equation I8-5 which can be written in terms of LRFD and ASD design as follows: LRFD

 Qut   t Qnt   

5/3

 Q    uv   v Qnv 

 15.6 kips     21.5 kips  0.96  1.0

ASD 5/3

5/3

5/3

 1.0 (from Spec. Eq. I8-5)

 10.4 kips     18.7 kips  o.k.

 Qat     Qnt t 

5/3

5/3

 0.96

5/3

 Qav     Qnv v 

 1.0 (from Spec. Eq. I8-5)

 10.5 kips   7.00 kips      14.4 kips    12.4 kips  0.98  1.0 o.k.

5/3

 0.98

Thus, a w-in.-diameter  6x-in.-long headed stud anchor is adequate for the applied loads. Limits of Application The application of the steel anchors in composite component provisions have strict limitations as summarized in the User Note provided at the beginning of AISC Specification Section I8.3. These provisions do not apply to typical composite beam designs nor do they apply to hybrid construction where the steel and concrete do not resist loads together via composite action such as in embed plates. This design example is intended solely to illustrate the calculations associated with an isolated anchor that is part of an applicable composite system. Available Strength Table Table I.12-1 provides available shear and tension strengths for standard Type B steel headed stud anchors conforming to the requirements of AWS D1.1 for use in composite components.

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Table I.12-1 Steel Headed Stud Anchor Available Strengths Anchor Shank Diameter

Asa

in. 2 s w d 1 ASD v = 2.31 t = 2.00

in.2 0.196 0.307 0.442 0.601 0.785 LRFD v = 0.65 t = 0.75

a

Qnv/v

vQnv

Qnv/v

vQnv

kips ASD 5.52 8.63 12.4 16.9 22.1

kips LRFD 8.30 13.0 18.7 25.4 33.2

kips ASD 6.38 9.97 14.4 N/Aa 25.5

kips LRFD 9.57 15.0 21.5 N/Aa 38.3

d-in.-diameter anchors conforming to AWS D1.1, Figure 7.1, do not meet the minimum head-to-shank diameter ratio of 1.6 as required for tensile resistance per AISC Specification Section I8.3.

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EXAMPLE I.13

COMPOSITE COLLECTOR BEAM DESIGN

Given: Determine if the composite beam designed in Example I.1 is adequate to serve as a collector beam for the transfer of wind-induced compression forces in combination with gravity loading as indicated in Figure I.13. Applied forces were generated from an elastic analysis and stability shall be accounted for using the effective length method of design.

Fig. I.13. Composite collector beam and applied loading elevation.

Solution: From AISC Manual Table 1-1, the geometric properties are as follows: W2150

A = 14.7 in.2 bf = 6.53 in. tw = 0.380 in.

Ix = 984 in.4 d = 20.8 in. bf/2tf = 6.10

Iy = 24.9 in.4 rx = 8.18 in. h/tw = 49.4

J = 1.14 in.4 ry = 1.30 in. ho = 20.3 in.

Refer to Example I.1 for additional information regarding strength and serviceability requirements associated with pre-composite and composite gravity load conditions. Required Compressive Strength From ASCE/SEI 7, Chapter 2, the required axial strength for the governing load combination, including wind, is: LRFD

ASD

Pu  1.2 D  1.0W  L  1.2  0 kips   1.0  0.556 kip/ft  45 ft   0 kips

Pa  D  0.75L  0.75  0.6W   0 kips  0.75  0 kips   0.75  0.6  0.556 kip/ft  45 ft 

 25.0 kips

 11.3 kips Available Compressive Strength (General) The collector element is conservatively treated as a bare steel member for the determination of available compressive strength as discussed in AISC Specification Commentary Section I7. The effective length factor, K, for a pin-ended member is taken as 1.0 in accordance with Table C-A-7.1. Potential limit states are flexural buckling about both the minor and major axes, and torsional buckling.

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Lateral movement is assumed to be braced by the composite slab, thus weak-axis flexural buckling will not govern by inspection as Lcy = (KL)y = 0. The member is slender for compression as indicated in AISC Manual Table 1-1, thus strong-axis flexural buckling strength is determined in accordance with AISC Specification Section E7 for members with slender elements for Lcx = (KL)x = 45.0 ft. The composite slab will prevent the member from twisting about its shear center, thus torsional buckling is not a valid limit state; however, constrained-axis torsional buckling may occur as discussed in AISC Specification Commentary Section E4 with Lcz = (KL)z = 1.0(45 ft) = 45.0 ft. Compute the available compressive strengths for the limit states of strong-axis flexural buckling and constrainedaxis torsional buckling to determine the controlling strength. Strong-Axis Flexural Buckling Calculate the critical stress about the strong axis, Fcrx, in accordance with AISC Specification Section E3 as directed by Specification Section E7 for members with slender elements. Lcx  45.0 ft 12 in./ft   rx 8.18 in.  66.0 4.71

E 29, 000 ksi  4.71 Fy 50 ksi  113  66.0; therefore, use AISC Specification Equation E3-2

Fex 



2 E  Lcx   r   x 

(Spec. Eq. E3-4)

2

2  29, 000 ksi 

 66.0 2

 65.7 ksi Fy   Fcrx   0.658 Fex  Fy     50 ksi     0.658 65.7 ksi   50 ksi       36.4 ksi

(Spec. Eq. E3-2)

Classify each component of the wide-flange member for local buckling. Flange local buckling classification as determined from AISC Specification Table B4.1a, Case 1:

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 r  0.56  0.56

E Fy 29, 000 ksi 50 ksi

 13.5 

bf 2t f

 6.10  13.5; therefore, the flanges are nonslender

Therefore, the flanges are fully effective. Web local buckling classification as determined from AISC Specification Table B4.1a, Case 5: E Fy

 r  1.49

29, 000 ksi 50 ksi

 1.49  35.9

h tw  49.4  35.9; therefore, the web is slender



To evaluate the impact of web slenderness on strong-axis flexural buckling, determine if a reduced effective web width, he, is required in accordance with AISC Specification Section E7.1 as follows: r

Fy Fcrx

 35.9

50 ksi 36.4 ksi

 42.1    49.4; therefore, use AISC Specification Equation E7-3 to determine he

The effective width imperfection adjustment factors, c1 and c2, are selected from AISC Specification Table E7.1, Case (a): c1  0.18 c2  1.31 2

   Fel   c2 r  Fy      35.9    1.31   49.4     45.3 ksi

(Spec. Eq. E7-5) 2

 50 ksi 

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h h    tw  tw    49.4  0.380 in.  18.8 in.  F  F he  h  1  c1 el  el Fcr  Fcr   45.3ksi  45.3 ksi  18.8 in. 1  0.18  36.4 ksi  36.4 ksi   16.8 in.

(from Spec. Eq. E7-3)

Calculate the effective area of the section: Ae  A  (h  he )tw  14.7 in.2  18.8 in.  16.8 in. 0.380 in.  13.9 in.2 Calculate the nominal compressive strength: Pnx  Fcrx Ae

(Spec. Eq. E7-1)



  36.4 ksi  13.9 in.2



 506 kips

Calculate the available compressive strength: LRFD

ASD

c  0.90

c  1.67

c Pn  0.90  506 kips 

Pn 506 kips  c 1.67  303 kips

 455 kips

Constrained-Axis Torsional Buckling Assuming the composite slab provides a lateral bracing point at the top flange of the beam, the constrained-axis buckling stress, Fez, can be determined using AISC Specification Commentary Equaation C-E4-1 as follows: The distance to bracing point from shear center along weak axis: d 2 20.8 in.  2  10.4 in.

a

The distance to bracing point from shear center along strong axis is:

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b0

ro2  rx2  ry2  a 2  b 2

(Spec. Eq. C-E4-3)

  8.18 in.  1.30 in.  10.4 in.   0 in. 2

2

2

2

 177 in.2

From AISC Specification Commentary Section E4, the finite brace stiffness factor is:   0.9

 2 EI y Fez      Lcz 2

 1  ho 2   a 2   GJ  2   Aro  4 



 2  29, 000 ksi  24.9 in.4   0.9  2   45.0 ft 12 in./ft    1   14.7 in.2 177 in.2   6.20 ksi







(Spec. Eq. C-E4-1)

   20.3 in. 

4

2

 2  10.4 in.   11, 200 ksi  1.14 in.4 









   

To evaluate the impact of web slenderness on constrained-axis torsional buckling, determine if a reduced effective web width, he, is required in accordance with AISC Specification Section E7.1 as follows: r

Fy Fcr

 35.9

50 ksi 6.20 ksi

 102    46.4; therefore use AISC Specification Equation E7-2 he  h

(from Spec. Eq. E7-2)

Thus the full steel area may be used without reduction and the available compressive strength for constrained axis buckling strength is calculated as follows: Lcz   KL  z

  45.0 ft 12 in./ft   540 in.

Fy 50 ksi  Fez 6.20 ksi  8.06  2.25, therefore, use AISC Specification Equation E3-3

Fcrz  0.877 Fez

(Spec. Eq. E3-3)

 0.877  6.20 ksi   5.44 ksi The nominal compressive strength is calculated with no reduction for slenderness, Ae = A, as follows:

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Pnz  Fcrz Ae

(Spec. Eq. E7-1)

  5.44 ksi  14.7 in.

2



 80.0 kips

The available compressive strength is determined as follows: LRFD

ASD

c  0.90

c  1.67

c Pnz  0.90  80.0 kips 

Pnz 80.0 kips  c 1.67  47.9 kips

 72.0 kips

Note that it may be possible to utilize the flexural stiffness and strength of the slab as a continuous torsional restraint, resulting in increased constrained-axis torsional buckling capacity; however, that exercise is beyond the scope of this design example. A summary of the available compressive strength for each of the viable limit states is as follows: LRFD

ASD

Strong-axis flexural buckling:

Strong-axis flexural buckling: Pnx  303 kips c

c Pnx  455 kips

Constrained-axis torsional buckling: c Pnz  72.0 kips

Constrained-axis torsional buckling: Pnz  47.9 kips controls c

controls

Required First-Order Flexural Strength From ASCE/SEI 7, Chapter 2, the required first-order flexural strength for the governing load combination including wind is: LRFD

ASD

wu  1.2 D  1.0W  L  1.2  0.9 kip/ft   1.0  0 kip/ft   1 kip/ft  2.08 kip/ft

Mu  

wa  D  0.75 L  0.75  0.6W   0.9 kip/ft  0.75 1 kip/ft   0.75  0.6  0 kip/ft   1.65 kip/ft

wu L2 8

Ma 

 2.08 kip/ft  45 ft 2



8  527 kip-ft

wa L2 8

1.65 kip/ft  45 ft 2

8  418 kip-ft

Required Second-Order Flexural Strength The effective length method is utilized to consider stability for this element as permitted by AISC Specification Section C1.2 and Appendix 7.2. The addition of axial load will magnify the required first-order flexural strength

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due to member slenderness (P-δ) effects. This magnification (second-order analysis) can be approximated utilizing the procedure provided in AISC Specification Appendix 8 as permitted by Section C2.1b. Calculate the elastic critical buckling strength of the member in the plane of bending (in this case about the strongaxis of the beam) from AISC Specification Appendix 8, Section 8.2.1. For the effective length method, EI* is taken as EI in accordance with Appendix 8.2.1, and the effective length, Lcx is taken as (KL)x in accordance with Appendix 7.2.3. As illustrated previously, K, is taken as 1.0 for a pin-ended member. Conservatively using the bare steel beam moment of inertia, the buckling strength is calculated as follows: Pe1  



2 EI *

(Spec. Eq. A-8-5)

 Lc1 2 2 EI

 KL 2x

(for the effective length method)



2  29, 000 ksi  984 in.4

 45.0 ft 12 in./ft    966 kips



2

For beam-columns subject to transverse loading between supports, the value of Cm is taken as 1.0 as permitted by AISC Specification Appendix 8, Section 8.2.1(b), and B1 is calculated from Specification Equation A-8-3 as follows: LRFD B1 

Cm 1 1   Pu Pe1

ASD B1 

1.0 1  25.0 kips  1  1.0    966 kips   1.03 

Cm 1 1   Pa Pe1

1.0 1  11.3 kips  1  1.6    966 kips   1.02 

Noting that the first-order moment is induced by vertical dead and live loading, it is classified as a non-translational moment, Mnt, in accordance with AISC Specification Section 8.2. The required second-order flexural strength is therefore calculated using AISC Specification Equation A-8-1 as: LRFD M u  B1 M nt  B2 M lt

ASD M a  B1 M nt  B2 M lt

 1.03  527 kip-ft   0

 1.02  418 kip-ft   0

 543 kip-ft

 426 kip-ft

Available Flexural Strength The available flexural strength of the composite beam is calculated in Example I.1 as: LRFD b M nx  769 kip-ft

ASD M nx  512 kip-ft b

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Interaction of Axial Force and Flexure Interaction between axial forces and flexure in composite collector beams is addressed in AISC Specification Commentary Section I7, which states that the non-composite axial strength and the composite flexural strength may be used with the interaction equations provided in Chapter H as a reasonable simplification for design purposes. This procedure is illustrated as follows: LRFD

ASD

c Pn  72.0 kips

Pn  47.9 kips c

b M nx  769 kip-ft

M nx  512 kip-ft c

Pr P = u Pc c Pn 25.0 kips  72.0 kips  0.347  0.2

Pr Pa  Pc Pn / c 11.3 kips  47.9 kips  0.236  0.2

Therefore, use AISC Specification Equation H1-1a.

Therefore, use AISC Specification Equation H1-1a.

Pu 8  Mu   c Pn 9  b M nx

Pa 8  Ma    1.0 Pn / c 9  M nx / b  8  426 kip-ft  0.236     1.0 9  512 kip-ft 

   1.0 

8  543 kip-ft  0.347     1.0 9  769 kip-ft  0.975  1.0 o.k.

0.976  1.0

o.k.

The collector element is adequate to resist the imposed loads. Load Introduction Effects AISC Specification Commentary Section I7 indicates that the effect of the vertical offset between the plane of the diaphragm and the collector element should be investigated. It has been shown that the resulting eccentricity between the plane of axial load introduction in the slab and the centroid of the beam connections does not result in any additional flexural demand assuming the axial load is introduced uniformly along the length of the beam; however, this eccentricity will result in additional shear reactions (Burmeister and Jacobs, 2008). The additional shear reaction assuming an eccentricity of d/2 is calculated as follows: LRFD Vu -add  

Pu d 2L  25.0 kips  20.8 in. 2  45 ft 12 in./ft 

 0.481 kips

ASD Va -add  

Pa d 2L 11.3kips  20.8 in. 2  45 ft 12 in./ft 

 0.218 kips

As can be seen from these results, the additional vertical shear due to the axial collector force is quite small and in most instances will be negligible versus the governing shear resulting from gravity-only load combinations.

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Shear Connection AISC Specification Commentary Section I7 notes that it is not required to superimpose the horizontal shear due to lateral forces with the horizontal shear due to flexure for the determination of steel anchor requirements, thus the summation of nominal strengths for all steel anchors along the beam length may be used for axial force transfer. Specific resistance and safety factors for this condition are not provided in Section I8.2 as they are implicitly accounted for within the system resistance and safety factors used for the determination of the available flexural strength of the beam. Until additional research becomes available, a conservative approach is to apply the composite component factors from Specification Section I8.3 to the nominal steel anchor strengths determined from Specification Section I8.2. From Example I.1, the strength for w-in.-diameter anchors in normal weight concrete with f c  4 ksi and deck oriented perpendicular to the beam is: 1 anchor per rib: 2 anchors per rib:

Qn  17.2 kips/anchor Qn  14.6 kips/anchor

Over the entire beam length, there are 42 anchors in positions with one anchor per rib and four anchors in positions with two anchors per rib, thus the total available strength for diaphragm shear transfer is: LRFD

ASD

v  0.65

 v  2.31

c Pn  0.65  42 17.2 kips/anchor   4(14.6 kips/anchor) 

Pn 42 17.2 kips/anchor   4 14.6 kips/anchor   c 2.31  338 kips  11.3 kips o.k.

 508 kips  25.0 kips

o.k.

Note that the longitudinal available shear strength of the diaphragm itself (consisting of the composite deck and concrete fill) will often limit the amount of force that can be introduced into the collector beam and should also be evaluated as part of the overall design. Summary A W2150 collector with 46, w-in.-diameter by 4d-in.-long, steel headed stud anchors is adequate to resist the imposed loads.

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CHAPTER I DESIGN EXAMPLE REFERENCES ACI 318 (2014), Building Code Requirements for Structural Concrete, ACI 318-14; and Commentary, ACI 318R14, American Concrete Institute, Farmington Hills, MI. ASCE (2014), Design Loads on Structures During Construction, ASCE/SEI 37-14, American Society of Civil Engineers, Reston, VA. AWS (2015), Structural Welding Code—Steel, AWS D1.1/D1.1M:2015, American Welding Society, Miami, FL. Burmeister, S. and Jacobs, W.P. (2008), “Under Foot: Horizontal Floor Diaphragm Load Effects on Composite Beam Design,” Modern Steel Construction, AISC, December. Griffis, L.G. (1992), Load and Resistance Factor Design of W-Shapes Encased in Concrete, Design Guide 6, AISC, Chicago, IL. ICC (2015), International Building Code, International Code Council, Falls Church, VA. Leon, R.T. and Hajjar, J.F. (2008), “Limit State Response of Composite Columns and Beam-Columns Part 2: Application of Design Provisions for the 2005 AISC Specification,” Engineering Journal, AISC, Vol. 45, No. 1, pp. 21–46. Murray, T.M., Allen, D.E., Ungar, E.E. and Davis, D.B. (2016), Floor Vibrations Due to Human Activity, Design Guide 11, 2nd Ed., AISC, Chicago, IL. Park, R. and Gamble, W.L. (2000), Reinforced Concrete Slabs, 2nd Ed., John Wiley & Sons, New York, NY. SDI (2011), Standard for Composite Steel Floor Deck-Slabs, ANSI/SDI C1.0-2011, Glenshaw, PA. West, M.A. and Fisher, J.M. (2003), Serviceability Design Consideration for Steel Buildings, Design Guide 3, 2nd Ed., AISC, Chicago, IL. Young, W.C. and Budynas, R.C. (2002), Roark’s Formulas for Stress and Strain, 7th Ed., McGraw-Hill, New York, NY.

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Chapter J Design of Connections AISC Specification Chapter J addresses the design of connections. The chapter’s primary focus is the design of welded and bolted connections. Design requirements for fillers, splices, column bases, concentrated forces, anchors rods and other threaded parts are also covered. See AISC Specification Appendix 3 for special requirements for connections subject to fatigue.

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EXAMPLE J.1

FILLET WELD IN LONGITUDINAL SHEAR

Given: As shown in Figure J.1-1, a ¼-in.-thick  18-in. wide plate is fillet welded to a a-in.-thick plate. The plates are ASTM A572 Grade 50 and have been properly sized. Use 70-ksi electrodes. Note that the plates could be specified as ASTM A36, but Fy = 50 ksi plate has been used here to demonstrate the requirements for long welds. Confirm that the size and length of the welds shown are adequate to resist the applied loading.

Fig. J.1-1. Geometry and loading for Example J.1. Solution: From AISC Manual Table 2-5, the material properties are as follows: ASTM A572 Grade 50 Fy = 50 ksi Fu = 65 ksi From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2  33 kips   1.6 100 kips 

 200 kips

ASD

Pa  33 kips  100 kips  133 kips

Maximum and Minimum Weld Size Because the thickness of the overlapping plate is ¼ in., the maximum fillet weld size that can be used without special notation per AISC Specification Section J2.2b, is a x-in. fillet weld. A x-in. fillet weld can be deposited in the flat or horizontal position in a single pass (true up to c-in.). From AISC Specification Table J2.4, the minimum size of the fillet weld, based on a material thickness of 4 in. is 8 in.

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Weld Strength The nominal weld strength per inch of x-in. weld, determined from AISC Specification Section J2.4(b) is:

Rn  Fnw Awe

(Spec. Eq. J2-4)

  0.60 FEXX  Awe  x in.   0.60  70 ksi     2   5.57 kip/in. From AISC Specification Section J2.2b, check the weld length to weld size ratio, because this is an end-loaded fillet weld. l 27.0 in.  w x in.  144  100; therefore, AISC Specification Equation J2-1 must be applied   1.2  0.002  l w   1.0

(Spec. Eq. J2-1)

 1.2  0.002 144   1.0  0.912

The nominal weld shear rupture strength is: Rn  0.912  5.57 kip/in. 2 welds  27 in.  274 kips From AISC Specification Section J2.4, the available shear rupture strength is: LRFD

ASD

  0.75 

  2.00

Rn = 0.75  274 kips 

Rn 274 kips =  2.00 = 137 kips  133 kips o.k.

= 206 kips > 200 kips

o.k.

The base metal strength is determined from AISC Specification Section J2.4(a). The 4-in.-thick plate controls:

Rn  FnBM ABM

(Spec. Eq. J2-2)

 0.60 Fu t p lweld  0.60  65 ksi 4 in. 2 welds  27 in.  527 kips LRFD   0.75 

Rn = 0.75  527 kips  = 395 kips > 200 kips

o.k.

ASD

  2.00   Rn 527 kips   2.00  264 kips  133 kips o.k.

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EXAMPLE J.2

FILLET WELD LOADED AT AN ANGLE

Given:

Verify a fillet weld at the edge of a gusset plate is adequate to resist a force of 50 kips due to dead load and 150 kips due to live load, at an angle of 60° relative to the weld, as shown in Figure J.2-1. Assume the beam and the gusset plate thickness and length have been properly sized. Use a 70-ksi electrode.

Fig. J.2-1. Geometry and loading for Example J.2. Solution:

From ASCE/SEI 7, Chapter 2, the required tensile strength is: LRFD Pu  1.2  50 kips   1.6 150 kips 

ASD

Pa  50 kips  150 kips  200 kips

 300 kips

Assume a c-in. fillet weld is used on each side of the plate. Note that from AISC Specification Table J2.4, the minimum size of fillet weld, based on a material thickness of w in. is 4 in. (assuming the beam flange thickness exceeds w in.). Available Shear Strength of the Fillet Weld Per Inch of Length From AISC Specification Section J2.4(b), the nominal strength of the fillet weld is determined as follows: Rn  Fnw Awe



(Spec. Eq. J2-4)



 0.60 FEXX 1.0  0.50sin1.5 60 Awe  c in.   0.60  70 ksi  1.0 + 0.50sin1.5 60    2   13.0 kip/in.





From AISC Specification Section J2.4(b), the available shear strength per inch of weld for fillet welds on two sides is:

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LRFD

ASD

  0.75

  2.00

Rn  0.75 13.0 kip/in. 2 sides 

Rn 13.0 kip/in.   2 sides   2.00  13.0 kip/in.

 19.5 kip/in. Required Length of Weld LRFD

ASD

300 kips l 19.5 kip/in.  15.4 in.

200 kips l 13.0 kip/in.  15.4 in.

Use 16 in. on each side of the plate.

Use 16 in. on each side of the plate.

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EXAMPLE J.3

COMBINED TENSION AND SHEAR IN BEARING-TYPE CONNECTIONS

Given:

A w-in.-diameter, Group A bolt with threads not excluded from the shear plane (thread condition N) is subjected to a tension force of 3.5 kips due to dead load and 12 kips due to live load, and a shear force of 1.33 kips due to dead load and 4 kips due to live load. Check the combined stresses according to AISC Specification Equations J3-3a and J3-3b. Solution:

From ASCE/SEI 7, Chapter 2, the required tensile and shear strengths are: LRFD Tension: Tu  1.2  3.5 kips   1.6 12 kips 

ASD Tension: Ta  3.5 kips  12 kips

 15.5 kips

 23.4 kips

Shear: Va  1.33kips  4 kips

Shear: Vu  1.2 1.33kips   1.6  4 kips 

 5.33 kips

 8.00 kips Available Tensile Strength

When a bolt is subject to combined tension and shear, the available tensile strength is determined according to the limit states of tension and shear rupture, from AISC Specification Section J3.7 as follows. From AISC Specification Table J3.2, Group A bolts: Fnt = 90 ksi Fnv = 54 ksi From AISC Manual Table 7-2, for a w-in.-diameter bolt: Ab = 0.442 in.2 The available shear stress is determined as follows and must equal or exceed the required shear stress. LRFD

ASD

  0.75

  2.00

Fnv  0.75  54 ksi 

Fnv 54 ksi   2.00  27.0 ksi

 40.5 ksi

f rv  

Vu Ab 8.00 kips

0.442 in.2  18.1 ksi  40.5 ksi o.k.

f rv  

Va Ab 5.33 kips

0.442 in.2  12.1 ksi  27.0 ksi o.k.

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The available tensile strength of a bolt subject to combined tension and shear is as follows: LRFD Fnt Fnt  1.3Fnt  f rv  Fnt (Spec. Eq. J3-3a) Fnv 90 ksi  1.3  90 ksi   18.1 ksi   90 ksi 40.5 ksi  76.8 ksi

ASD Fnt Fnt  1.3Fnt  f rv  Fnt (Spec. Eq. J3-3b) Fnv 90 ksi  1.3  90 ksi   12.1 ksi   90 ksi 27.0 ksi  76.7 ksi

For combined tension and shear,   0.75, from AISC Specification Section J3.7.

For combined tension and shear,   2.00, from AISC Specification Section J3.7.

Rn  Fnt Ab

Rn Fnt Ab   



 0.75  76.8 ksi  0.442 in.  25.5 kips  23.4 kips

2



o.k.

(Spec. Eq. J3-2)



(Spec. Eq. J3-2)

 76.7 ksi   0.442 in.2 

2.00  17.0 kips  15.5 kips o.k.

The effects of combined shear and tensile stresses need not be investigated if either the required shear or tensile stress is less than or equal to 30% of the corresponding available stress per the User Note at the end of AISC Specification Section J3.7. In the example herein, both the required shear and tensile stresses exceeded the 30% threshold and evaluation of combined stresses was necessary. AISC Specification Equations J3-3a and J3-3b may be rewritten so as to find a nominal shear stress, Fnv , as a function of the required tensile stress as is shown in AISC Specification Commentary Equations C-J3-7a and C-J37b.

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EXAMPLE J.4A SLIP-CRITICAL CONNECTION WITH SHORT-SLOTTED HOLES Slip-critical connections shall be designed to prevent slip and for the limit states of bearing-type connections.

Given: Refer to Figure J.4A-1 and select the number of bolts that are required to support the loads shown when the connection plates have short slots transverse to the load and no fillers are provided. Select the number of bolts required for slip resistance only.

Fig. J.4A-1. Geometry and loading for Example J.4A. Solution: From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2 17 kips   1.6  51 kips   102 kips

ASD

Pa  17 kips  51 kips  68.0 kips

From AISC Specification Section J3.8(a), the available slip resistance for the limit state of slip for standard size and short-slotted holes perpendicular to the direction of the load is determined as follows:    = 1.00   = 1.50  = 0.30 for Class A surface Du = 1.13 hf = 1.0, no filler is provided Tb = 28 kips, from AISC Specification Table J3.1, Group A ns = 2, number of slip planes

Rn  Du h f Tb ns

(Spec. Eq. J3-4)

 0.30 1.131.0  28 kips  2   19.0 kips/bolt The available slip resistance is:

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LRFD Rn  1.00 19.0 kips/bolt   19.0 kips/bolt

ASD Rn 19.0 kips/bolt   1.50  12.7 kips/bolt

Required Number of Bolts LRFD

ASD

P nb  u Rn 102 kips  19.0 kips/bolt  5.37 bolts

P nb  a  Rn     68.0 kips  12.7 kips/bolt  5.35 bolts

Use 6 bolts

Use 6 bolts

Note: To complete the verification of this connection, the limit states of bolt shear, bearing, tearout, tensile yielding, tensile rupture, and block shear rupture must also be checked.

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EXAMPLE J.4B SLIP-CRITICAL CONNECTION WITH LONG-SLOTTED HOLES Given: Repeat Example J.4A with the same loads, but assuming that the connection plates have long-slotted holes in the direction of the load, as shown in Figure J.4B-1.

Fig. J.4B-1. Geometry and loading for Example J.4B.

Solution: The required strength from Example J.4A is: LRFD

Pu  102 kips

ASD

Pa  68.0 kips

From AISC Specification Section J3.8(c), the available slip resistance for the limit state of slip for long-slotted holes is determined as follows:    = 0.70   = 2.14  = 0.30 for Class A surface Du = 1.13 hf = 1.0, no filler is provided Tb = 28 kips, from AISC Specification Table J3.1, Group A ns = 2, number of slip planes

Rn  Du h f Tb ns

(Spec. Eq. J3-4)

 0.30 1.131.0  28 kips  2   19.0 kips/bolt The available slip resistance is: LRFD Rn  0.70 19.0 kips/bolt   13.3 kips/bolt

ASD Rn 19.0 kips/bolt   2.14  8.88 kips/bolt

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Required Number of Bolts LRFD

ASD

P nb  u Rn 102 kips  13.3 kips/bolt  7.67 bolts

P nb  a R  n    68.0 kips  8.88 kips/bolt  7.66 bolts

Use 8 bolts

Use 8 bolts

Note: To complete the verification of this connection, the limit states of bolt shear, bearing, tearout, tensile yielding, tensile rupture, and block shear rupture must be determined.

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EXAMPLE J.5

COMBINED TENSION AND SHEAR IN A SLIP-CRITICAL CONNECTION

Because the pretension of a bolt in a slip-critical connection is used to create the clamping force that produces the shear strength of the connection, the available shear strength must be reduced for any load that produces tension in the connection.

Given: The slip-critical bolt group shown in Figure J.5-1 is subjected to tension and shear. This example shows the design for bolt slip resistance only, and assumes that the beams and plates are adequate to transmit the loads. Determine if the bolts are adequate.

Fig. J.5-1. Geometry and loading for Example J.5.

Solution: From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2 15 kips   1.6  45 kips   90.0 kips By geometry:

ASD

Pa  15 kips  45 kips  60.0 kips By geometry: 4  60.0 kips  5  48.0 kips

4  90.0 kips  5  72.0 kips

Ta 

3  90.0 kips  5  54.0 kips

Va 

Tu 

Vu 

3  60.0 kips  5  36.0 kips

Available Bolt Tensile Strength The available tensile strength is determined from AISC Specification Section J3.6. From AISC Specification Table J3.2 for Group A bolts, the nominal tensile strength in ksi is, Fnt = 90 ksi. From AISC Manual Table 7-1, for a w-in.-diameter bolt:

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Ab  0.442 in.2 The nominal tensile strength is: Rn  Fnt Ab



  90 ksi  0.442 in.

2

(from Spec. Eq. J3-1)



 39.8 kips

The available tensile strength is:   0.75 

LRFD

  2.00 

72.0 kips  8 bolts  29.9 kips/bolt  9.00 kips/bolt o.k.

Rn  0.75  39.8 kips/bolt  

ASD

Rn 39.8 kips/bolt 48.0 kips    2.00 8 bolts  19.9 kips/bolt  6.00 kips/bolt

o.k.

Note that the available tensile strength per bolt can also be taken from AISC Manual Table 7-2. Available Slip Resistance per Bolt The available slip resistance for one bolt in standard size holes is determined using AISC Specification Section J3.8(a):   = 1.00   = 1.50  = 0.30 for Class A surface Du = 1.13 hf = 1.0, factor for fillers, assuming no more than one filler Tb = 28 kips, from AISC Specification Table J3.1, Group A ns = 1, number of slip planes LRFD Determine the available slip resistance (Tu = 0) of a bolt:

ASD Determine the available slip resistance (Ta = 0) of a bolt:

Rn  Du h f Tb ns

Rn Du h f Tb ns  (from Spec. Eq. J3-4)   0.30 1.131.0  28 kips 1 = 1.50  6.33 kips/bolt

(from Spec. Eq. J3-4)

 1.00  0.30 1.131.0  28 kips 1  9.49 kips/bolt

Note that the available slip resistance for one bolt with a Class A faying surface can also be taken from AISC Manual Table 7-3. Available Slip Resistance of the Connection Because the slip-critical connection is subject to combined tension and shear, the available slip resistance is multiplied by a reduction factor provided in AISC Specification Section J3.9.

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LRFD Slip-critical combined tension and shear factor:

Tu 0 DuTb nb 72.0 kips  1 0 1.13  28 kips  8 

ksc  1 

(Spec. Eq. J3-5a)

ksc  1 

 1

 0.716  Rn = Rn k sc nb

1.5Ta 0 DuTb nb

1.5  48.0 kips 

1.13  28 kips  8

(Spec. Eq. J3-5b)

0

 0.716

  9.49 kips/bolt  0.716  8 bolts   54.4 kips  54.0 kips o.k.

ASD Slip-critical combined tension and shear factor:

Rn R = n k sc nb     6.33 kips/bolt  0.716  8 bolts   36.3 kips  36.0 kips o.k.

Note: The bolt group must still be checked for all applicable strength limit states for a bearing-type connection.

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EXAMPLE J.6

BASE PLATE BEARING ON CONCRETE

Given: As shown in Figure J.6-1, an ASTM A992 column bears on a concrete pedestal with fc = 3 ksi. The space between the base plate and the concrete pedestal has grout with fc = 4 ksi. Verify the ASTM A36 base plate will support the following loads in axial compression: PD = 115 kips PL = 345 kips

Fig. J.6-1. Geometry for Example J.6.

Solution: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Column ASTM A992 Fy = 50 ksi Fu = 65 ksi Base Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Column W1296 d = 12.7 in. bf = 12.2 in. tf = 0.900 in. tw = 0.550 in.

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From ASCE/SEI 7, Chapter 2, the required compressive strength is: LRFD Pu  1.2 115 kips   1.6  345 kips 

ASD

Pa  115 kips  345 kips

 460 kips

 690 kips Base Plate Dimensions

Determine the required base plate area from AISC Specification Section J8 conservatively assuming bearing on the full area of the concrete support. LRFD

ASD

c  0.65   A1 req  



Pu c 0.85 f c

(from Spec. Eq. J8-1)

690 kips 0.65  0.85 3 ksi 

c  2.31    P A1 req   c a 0.85 f c



(from Spec. Eq. J8-1)

2.31 460 kips  0.85  3 ksi 

 417 in.2

 416 in.2

Note: The strength of the grout has conservatively been neglected, as its strength is greater than that of the concrete pedestal. Try a 22-in.  22-in. base plate. Verify N  d  2  3 in. and B  b f  2  3 in. for anchor rod pattern shown in diagram: d  2  3 in.  12.7 in.  2  3 in.  18.7 in.  22 in. o.k.

b f  2  3 in.  12.2 in.  2  3 in.

 18.2 in.  22 in. o.k. Base plate area:

A1  NB   22 in. 22 in.  484 in.2  417 in.2

o.k. (conservatively compared to ASD value for A1( req ) )

Note: A square base plate with a square anchor rod pattern will be used to minimize the chance for field and shop problems. Concrete Bearing Strength Use AISC Specification Equation J8-2 because the base plate covers less than the full area of the concrete support. Because the pedestal is square and the base plate is a concentrically located square, the full pedestal area is also the geometrically similar area. Therefore:

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A2   24 in. 24 in.  576 in.2 The available bearing strength is: LRFD

ASD c  2.31

c  0.65 c Pp  c 0.85 f c A1

A2

Pp 0.85 f c A1  c c

 c 1.7 f c A1

A1

(from Spec. Eq. J8-2)



 0.65  0.85  3 ksi  484 in.2





2

576 in.

484 in.2

 0.65 1.7  3 ksi  484 in.2



 875 kips  1, 600 kips, use 875 kips

875 kips > 690 kips o.k. 



A2 A1





1.7 f c A1 c

0.85  3 ksi  484 in. 2.31 

2





(from Spec. Eq. J8-2)

576 in.2 484 in.2

1.7  3 ksi  484 in.2



2.31  583 kips  1, 070 kips, use 583 kips

583 kips > 460 kips o.k. 

Notes: 1. A2 A1  4; therefore, the upper limit in AISC Specification Equation J8-2 does not control. 2. As the area of the base plate approaches the area of concrete, the modifying ratio, A2 A1 , approaches unity and AISC Specification Equation J8-2 converges to AISC Specification Equation J8-1. Required Base Plate Thickness

The base plate thickness is determined in accordance with AISC Manual Part 14. m



N  0.95d 2 22 in.  0.95 12.7 in.

(Manual Eq. 14-2)

2

 4.97 in. n



B  0.8b f

(Manual Eq. 14-3)

2 22 in.  0.8 12.2 in. 2

 6.12 in. n  

db f

(Manual Eq. 14-4)

4

12.7 in.12.2 in. 4

 3.11 in.

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LRFD  4db f X  d  bf 



2

 P  u  c Pp 

ASD (Manual Eq. 14-6a)

 4 12.7 in.12.2 in.   690 kips      12.7 in.  12.2 in.2   875 kips   0.788

 4db f X  d  bf 



2

 P  c a  Pp 

(Manual Eq. 14-6b)

 4 12.7 in.12.2 in.   460 kips      12.7 in.  12.2 in.2   583 kips   0.789

Conservatively, use the LRFD value for X. 

2 X 1 1 X

1

(Manual Eq. 14-5)

2 0.788



1 1  1  0.788  1.22  1, use   1

Note:  can always be conservatively taken equal to 1.

n  1 3.11 in.  3.11 in. l  max m, n, n  max 4.97 in., 6.12 in., 3.11 in.  6.12 in. LRFD f pu

ASD

P  u BN 

f pa 690 kips



 22 in. 22 in.

 1.43 ksi

From AISC Manual Equation 14-7b:

2 f pu

tmin  l

0.90 Fy

  6.12 in.

460 kips

 22 in. 22 in.

 0.950 ksi

From AISC Manual Equation 14-7a:

tmin  l

P  a BN

2 1.43 ksi 

1.67  2 f pa  Fy

  6.12 in.

0.90  36 ksi 

 1.82 in.

 1.82 in. Use PL2 in. 22 in. 1 ft 10 in., ASTM A36.

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Chapter K Additional Requirements for HSS and Box Section Connections Examples K.1 through K.6 illustrate common beam-to-column shear connections that have been adapted for use with HSS columns. Example K.7 illustrates a through-plate shear connection, which is unique to HSS columns. Calculations for transverse and longitudinal forces applied to HSS are illustrated in Example K.8. Examples of HSS base plate and end plate connections are given in Examples K.9 and K.10.

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EXAMPLE K.1

WELDED/BOLTED WIDE TEE CONNECTION TO AN HSS COLUMN

Given: Verify a connection between an ASTM A992 W1650 beam and an ASTM A500, Grade C, HSS884 column using an ASTM A992 WT-shape, as shown in Figure K.1-1. Design, assuming a flexible support condition, for the following vertical shear loads: PD = 6.2 kips PL = 18.5 kips Note: A tee with a flange width wider than 8 in. was selected to provide sufficient surface for flare bevel groove welds on both sides of the column, because the tee will be slightly offset from the column centerline.

Fig K.1-1. Connection geometry for Example K.1. Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Tee ASTM A992 Fy = 50 ksi Fu = 65 ksi Column ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi

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From AISC Manual Tables 1-1, 1-8 and 1-12, the geometric properties are as follows: W1650 tw = 0.380 in. d = 16.3 in. tf = 0.630 in. T = 13s in. WT524.5

tsw = tw = 0.340 in. d = 4.99 in. tf = 0.560 in. bf = 10.0 in. k1 = m in. (see W1049) HSS884 t = 0.233 in. B = 8.00 in.

From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2  6.2 kips   1.6 18.5 kips   37.0 kips

ASD Pa  6.2 kips  18.5 kips  24.7 kips

Calculate the available strength assuming a flexible support condition. Required Number of Bolts The required number of bolts will ultimately be determined using the coefficient, C, from AISC Manual Table 7-6. First, the available strength per bolt must be determined. Determine the available shear strength of a single bolt. From AISC Manual Table 7-1, for w-in.-diameter Group A bolts: LRFD

ASD rn  11.9 kips 

rn  17.9 kips

The edge distance is checked against the minimum edge distance requirement provided in AISC Specification Table J3.4. lev  14 in.  1 in.

o.k.

The available bearing and tearout strength per bolt on the tee stem based on edge distance is determined from AISC Manual Table 7-5, for lev = 14 in., as follows: LRFD rn   49.4 kip/in. 0.340 in.  16.8 kips

ASD rn   32.9 kip/in. 0.340 in.   11.2 kips

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The bolt spacing is checked against the minimum spacing requirement between centers of standard holes provided in AISC Specification Section J3.3. 2qd  2q  w in.  2.00 in.  s  3 in.

o.k.

The available bearing and tearout strength per bolt on the tee stem based on spacing is determined from AISC Manual Table 7-4, for s = 3 in., as follows: LRFD rn   87.8 kip/in. 0.340 in.

ASD rn   58.5 kip/in. 0.340 in.   19.9 kips

 29.9 kips

Bolt bearing and tearout strength based on edge distance controls over the available shear strength of the bolt. Determine the coefficient for the eccentrically loaded bolt group. LRFD

Cmin

ASD

P  u rn 37.0 kips  16.8 kips  2.20

Cmin

P  a rn /  24.7 kips  11.2 kips  2.21

Using e = 3 in. and s = 3 in., determine C from AISC Manual Table 7-6, Angle = 0.

Using e = 3 in. and s = 3 in., determine C from AISC Manual Table 7-6, Angle = 0.

Try four rows of bolts:

Try four rows of bolts:

C  2.81  2.20 o.k.

C  2.81  2.21 o.k.

Tee Stem Thickness and Length AISC Manual Part 9 stipulates a maximum tee stem thickness that should be provided for rotational ductility as follows: d  z in. 2 w in.   z in. 2  0.438 in.  0.340 in. o.k.

tsw max 

(from Manual Eq. 9-39)

Note: The beam web thickness is greater than the tee stem thickness. If the beam web were thinner than the tee stem, this check could be satisfied by checking the thickness of the beam web. As discussed in AISC Manual Part 10, it is recommended that the minimum length of a simple shear connection is one-half the T-dimension of the beam to be supported. The minimum length of the tee is determined as follow:

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T 2 13s in.  2  6.81 in.

lmin 

As discussed in AISC Manual Part 10, the detailed length of connection elements must be compatible with the Tdimension of the beam. The tee length is checked using the number of bolts, bolt spacing, and edge distances determined previously. l  3  3 in.  2 14 in.  11.5 in.  T  13s in. o.k.

Try l = 11.5 in. Tee Stem Shear Yielding Strength Determine the available shear strength of the tee stem based on the limit state of shear yielding from AISC Specification Section J4.2(a). Agv  lts  11.5 in. 0.340 in.  3.91 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  3.91 in.2



 117 kips

  1.00 

LRFD

ASD



 Rn  1.00 117 kips   117 kips  37.0 kips

  1.50 

o.k.

Rn 117 kips   1.50  78.0 kips  24.7 kips o.k.



Because of the geometry of the tee and because the tee flange is thicker than the stem and carries only half of the beam reaction, flexural yielding and shear yielding of the flange are not controlling limit states. Tee Stem Shear Rupture Strength Determine the available shear strength of the tee stem based on the limit state of shear rupture from AISC Specification Section J4.2(b). Anv  l  n  d n  z in.  ts  11.5 in.   4 m in.  z in.   0.340 in.  2.72 in.2

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Rn  0.60 Fu Anv



 0.60  65 ksi  2.72 in.2

(Spec. Eq. J4-4)



 106 kips

  0.75 

LRFD

  2.00  

 Rn  0.75 106 kips   79.5 kips  37.0 kips

ASD

Rn 106 kips   2.00  53.0 kips  24.7 kips o.k.

o.k.

Tee Stem Block Shear Rupture Strength The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3. Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the tee stem is determined as follows, using AISC Manual Tables 9-3a, 9-3b and 9-3c and AISC Specification Equation J4-5, with n = 4, leh = 1.99 in. (assume leh = 2.00 in. to use Table 93a), lev = 14 in. and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a:  F A  u nt  76.2 kip/in.  t  Shear yielding component from AISC Manual Table 9-3b:  0.60 Fy Agv  231 kip/in.   t

ASD Tension rupture component from AISC Manual Table 9-3a:





Fu Ant  50.8 kip/in.  t

 Shear yielding component from AISC Manual Table 9-3b:





0.60 Fy Agv  154 kip/in.  t





Shear rupture component from AISC Manual Table 9-3c:

Shear rupture component from AISC Manual Table 9-3c:

0.60 Fu Anv  210 kip/in.  t







0.60 Fu Anv  140 kip/in. t

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LRFD The design block shear rupture strength is:

ASD The allowable block shear rupture strength is:

Rn  0.60 Fu Avn  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant



Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant  +    140 kip/in.  50.8 kip/in. 0.340 in.

  210 kip/in.  76.2 kip/in. 0.340 in.   231 kip/in.  76.2 kip/in. 0.340 in.  97.3 kips  104 kips  97.3 kips  37.0 kips

 154 kip/in.  50.8 kip/in. 0.340 in.

o.k.

 64.9 kips  69.6 kips  64.9 kips  24.7 kips o.k.

Tee Stem Flexural Strength The required flexural strength for the tee stem is: LRFD

ASD

M u  Pu e

M a  Pa e

  37.0 kips  3 in.

  24.7 kips  3 in.

 111 kip-in.

 74.1 kip-in.

The tee stem available flexural strength due to yielding is determined as follows, from AISC Specification Section F11.1. The stem, in this case, is treated as a rectangular bar. Z  

ts d 2 4

 0.340 in.11.5 in.2 4 3

 11.2 in. Sx  

ts d 2 6

 0.340 in.11.5 in.2 6 3

 7.49 in.

M n  M p  Fy Z  1.6 Fy S x





(Spec. Eq. F11-1)



  50 ksi  11.2 in.3  1.6  50 ksi  7.49 in.3



 560 kip-in.  599 kip-in.  560 kips-in. Note: The 1.6 limit will never control for a plate because the shape factor (Z/S) for a plate is 1.5. The tee stem available flexural yielding strength is:

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LRFD

  0.90 

  1.67

 M n  0.90  560 kip-in.  504 kip-in.  111 kip-in.

ASD

M n 560 kip-in.  1.67   335 kip-in.  74.1 kip-in.

o.k.

o.k.

The tee stem available flexural strength due to lateral-torsional buckling is determined from Section F11.2. Lb d ts2



 3 in.11.5 in.  0.340 in.2

 298 0.08 E 0.08  29, 000 ksi   50 ksi Fy  46.4

1.9 E 1.9  29, 000 ksi   Fy 50 ksi  1,102 Because 46.4 < 298 < 1,102, Equation F11-2 is applicable with Cb = 1.00.   L d  Fy  M n  Cb 1.52  0.274  b2   M y  M p  t  E 

(Spec. Eq. F11-2)

  50 ksi   2 3  1.00 1.52  0.274  298      50 ksi  7.49in.   50 ksi  11.2in.  29, 000 ksi     517 kip-in.  560 kip-in.









 517 kip-in. LRFD

  0.90 

  1.67

 M n  0.90  517 kip-in.  465 kip-in.  111 kip-in.

ASD

M n 517 kip-in.  1.67   310 kip-in.  74.1 kip-in.

o.k.

o.k.

The tee stem available flexural rupture strength is determined from AISC Manual Part 9 as follows: Z net  

td 2  2tsw  d h  z in.1.5 in.  4.5 in. 4

 0.340 in.11.5 in.2 4

 2  0.340 in.m in.  z in.1.5 in.  4.5 in.

 7.67 in.3

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M n  Fu Z net



  65 ksi  7.67 in.3

(Manual Eq. 9-4)



 499 kip-in.

LRFD

ASD b  2.00

b  0.75  

M n  0.75  499 kip-in.  374 kip-in.  111 kip-in.

M n 499 kip-in.  2.00   250 kip-in.  74.1 kip-in.

o.k.

o.k.

Beam Web Bearing Because tw = 0.380 in. > tsw = 0.340 in., bolt bearing does not control the strength of the beam web. Weld Size Because the flange width of the tee is larger than the width of the HSS, a flare bevel groove weld is required. Taking the outside radius as R = 2t = 2(0.233 in.) = 0.466 in. and using AISC Specification Table J2.2, the effective throat thickness of the flare bevel groove weld is E = cR = c(0.466 in.) = 0.146 in. This effective throat thickness will be used for subsequent calculations; however, for the detail drawing, a x-in. weld is specified. Using AISC Specification Table J2.3, the minimum effective throat thickness of the flare bevel groove weld, based on the 0.233 in. thickness of the HSS column, is 8 in. E  0.146 in.  8 in.

The equivalent fillet weld that provides the same throat dimension is:  D  1      0.146  16   2  D  16 2  0.146   3.30 sixteenths of an inch

The equivalent fillet weld size is used in the following calculations. Weld Ductility Check weld ductility using AISC Manual Part 9. Let bf = B = 8.00 in.

b 

b f  2k1 2 8.00 in.  2 m in. 2

 3.19 in

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wmin  0.0155

 Fy t f 2  b 2  2  2    s  tsw b l 

(Manual Eq. 9-37)

 50 ksi  0.560 in.2   3.19 in.2   0.0155  2   s  0.340 in. 3.19 in.  11.5 in.2   0.158 in.  0.213in.

0.158 in. = 2.53 sixteenths of an inch Dmin  2.53  3.30 sixteenths of an inch

o.k.

Nominal Weld Shear Strength The load is assumed to act concentrically with the weld group (i.e., a flexible support condition). a = 0 and k = 0; therefore, C = 3.71 from AISC Manual Table 8-4, Angle = 0°.

Rn  CC1 Dl  3.711.00  3.30 sixteenths of an inch 11.5 in.  141 kips Shear Rupture of the HSS at the Weld tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  3.30 sixteenths 

62 ksi  0.164 in.  0.233 in.

By inspection, shear rupture of the tee flange at the welds will not control. Therefore, the weld controls. Available Weld Shear Strength From AISC Specification Section J2.4, the available weld strength is:   0.75 

LRFD

ASD



 Rn  0.75 141 kips   106 kips  37.0 kips

  2.00

o.k.



Rn 141 kips   2.00  70.5 kips  24.7 kips o.k.

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EXAMPLE K.2

WELDED/BOLTED NARROW TEE CONNECTION TO AN HSS COLUMN

Given:

Verify a connection for an ASTM A992 W1650 beam to an ASTM A500 Grade C HSS884 column using an ASTM A992 WT524.5 with fillet welds against the flat width of the HSS, as shown in Figure K.2-1. Use 70-ksi weld electrodes. Assume that, for architectural purposes, the flanges of the WT from the previous example have been stripped down to a width of 5 in. Design assuming a flexible support condition for the following vertical shear loads: PD = 6.2 kips PL = 18.5 kips Note: This is the same problem as Example K.1 with the exception that a narrow tee will be selected which will permit fillet welds on the flat of the column. The beam will still be centered on the column centerline; therefore, the tee will be slightly offset.

Fig K.2-1. Connection geometry for Example K.2. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Tee ASTM A992 Fy = 50 ksi Fu = 65 ksi Column ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi

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From AISC Manual Tables 1-1, 1-8 and 1-12, the geometric properties are as follows: W1650 tw = 0.380 in. d = 16.3 in. tf = 0.630 in. HSS884 t = 0.233 in. B = 8.00 in. WT524.5

tsw d tf k1

= tw = 0.340 in. = 4.99 in. = 0.560 in. = m in. (see W1049)

From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2  6.2 kips   1.6 18.5 kips 

ASD Pa  6.2 kips  18.5 kips  24.7 kips

 37.0 kips

The tee stem thickness, tee length, tee stem strength, and beam web bearing strength are verified in Example K.1. The required number of bolts is also determined in Example K.1. Maximum Tee Flange Width Assume 4-in. welds and HSS corner radius equal to 2.25 times the nominal thickness 2.25(4 in.) = b in. (refer to AISC Manual Part 1 discussion). The recommended minimum shelf dimension for 4-in. fillet welds from AISC Manual Figure 8-13 is 2 in. Connection offset (centerline of the column to the centerline of the tee stem): 0.380 in. 0.340 in. + = 0.360 in. 2 2

The stripped flange must not exceed the flat face of the tube minus the shelf dimension on each side: b f  8.00 in.  2  b in.  2 2 in.  2  0.360 in. 5.00 in.  5.16 in. o.k.

Minimum Fillet Weld Size From AISC Specification Table J2.4, the minimum fillet weld size = 8 in. (D = 2) for welding to 0.233-in.-thick material. Weld Ductility The flexible width of the connecting element, b, is defined in Figure 9-6 of AISC Manual Part 9:

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b 

b f  2k1 2 5.00 in.  2 m in. 2

 1.69 in. Fy t f 2 b

 b2   2  2    s  tsw l   50 ksi  0.560 in.2  1.69 in.2   0.0155  2   s  0.340 in. 1.69 in.  11.5 in.2   0.291 in.  0.213 in.; therefore, use wmin  0.213 in.

wmin  0.0155

(Manual Eq. 9-37)

Dmin   0.213 in.16   3.41 sixteenths of an inch

Try a 4-in. fillet weld as a practical minimum, which is less than the maximum permitted weld size of tf – z in. = 0.560 in. – z in. = 0.498 in., in accordance with AISC Specification Section J2.2b. Provide 2-in. return welds at the top of the tee to meet the criteria listed in AISC Specification Section J2.2b. Minimum HSS Wall Thickness to Match Weld Strength tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  4 

62 ksi  0.199 in.  0.233 in.

By inspection, shear rupture of the flange of the tee at the welds will not control. Therefore, the weld controls. Available Weld Shear Strength The load is assumed to act concentrically with the weld group (i.e., a flexible support condition). a = 0 and k = 0, therefore, C = 3.71 from AISC Manual Table 8-4, Angle = 0°.

Rn  CC1 Dl  3.711.00  4 sixteenths of an inch 11.5 in.  171 kips From AISC Specification Section J2.4, the available fillet weld shear strength is:

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  0.75 

LRFD

 Rn  0.75 171 kips   128 kips  37.0 kips

  2.00 

ASD

Rn 171 kips  2.00   85.5 kips  24.7 kips

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EXAMPLE K.3

DOUBLE-ANGLE CONNECTION TO AN HSS COLUMN

Given: Use AISC Manual Tables 10-1 and 10-2 to design a double-angle connection for an ASTM A992 W36231 beam to an ASTM A500 Grade C HSS14142 column, as shown in Figure K.3-1. The angles are ASTM A36 material. Use 70-ksi weld electrodes. The bottom flange cope is required for erection. Use the following vertical shear loads: PD = 37.5 kips PL = 113 kips

Fig K.3-1. Connection geometry for Example K.3. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Column ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi

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Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-1 and 1-12, the geometric properties are as follows: W36231 tw = 0.760 in. T = 31a in. HSS14142

t = 0.465 in. B = 14.0 in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  37.5 kips   1.6 113 kips 

ASD Ra  37.5 kips  113 kips  151 kips

 226 kips

Bolt and Weld Design Try eight rows of bolts and c-in. welds. Obtain the bolt group and angle available strength from AISC Manual Table 10-1, Group A. LRFD Rn  284 kips  226 kips

ASD Rn  189 kips  151 kips 

o.k. 

o.k.

Obtain the available weld strength from AISC Manual Table 10-2 (welds B). LRFD Rn  279 kips  226 kips

ASD Rn  186 kips  151 kips 

o.k. 

o.k.

Minimum Support Thickness The minimum required support thickness using AISC Manual Table 10-2 is determined as follows for Fu = 62 ksi material.  65 ksi  0.238 in.   = 0.250 in.  0.465 in.  62 ksi 

o.k.

Minimum Angle Thickness tmin  w  z in., from AISC Specification Section J2.2b  c in.  z in.  a in.

Use a-in. angle thickness to accommodate the welded legs of the double-angle connection.

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Use 2L432a1-112. Minimum Angle Length As discussed in AISC Manual Part 10, it is recommended that the minimum length of a simple shear connection is one-half the T-dimension of the beam to be supported. The minimum length of the connection is determined as follow: T 2 31a in.  2  15.7 in.  23.5 in. o.k.

lmin 

Minimum Column Width The workable flat for the HSS column is 11w in. from AISC Manual Table 1-12. The recommended minimum shelf dimension for c-in. fillet welds from AISC Manual Figure 8-13 is b in. The minimum acceptable width to accommodate the connection is: 2  4.00 in.   0.760 in.  2  b in.   9.89 in.  11w in.

o.k.

Available Beam Web Strength The available beam web strength, from AISC Manual design table discussion for Table 10-1, is the lesser of the limit states of block shear rupture, shear yielding, shear rupture, and the sum of the effective strengths of the individual fasteners. The beam is not coped, so the only applicable limit state is the effective strength of the individual fasteners. The effective strength of an individual fastener is the lesser of the fastener shear strength, bearing strength at the bolt hole, and the tearout strength at the bolt hole. For the limit state of fastener shear strength, with Ab = 0.442 in.2 from AISC Manual Table 7-1 for a w-in. bolt: rn  Fnv Ab



  54 ksi  0.442 in.2

  2 shear planes 

(from Spec. Eq. J3-1)

 47.7 kips/bolt

where Fnv is the nominal shear strength from AISC Specification Table J3.2 of a Group A bolt in a bearing-type connection when threads are not excluded from the shear planes. Assume that deformation at the bolt hole at service load is a design consideration. For the limit state of bearing: rn  2.4dtFu

(from Spec. Eq. J3-6a)

 2.4  w in. 0.760 in. 65 ksi   88.9 kips/bolt For the limit state of tearout:

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rn  1.2lc tFu

(from Spec. Eq. J3-6c)

 1.2  3 in.  m in. 0.760 in. 65 ksi   130 kips/bolt

where lc is the clear distance, in the direction of the force, between the edges of the bolt holes. Fastener shear strength is the governing limit state for all bolts at the beam web. Fastener shear strength is one of the limit states included in the available strength given in Table 10-1 and was previously shown to be adequate.

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EXAMPLE K.4

UNSTIFFENED SEATED CONNECTION TO AN HSS COLUMN

Given:

Use AISC Manual Table 10-6 to verify an unstiffened seated connection for an ASTM A992 W2162 beam to an ASTM A500 Grade C HSS12122 column, as shown in Figure K.4-1. The angles are ASTM A36 material. Use 70-ksi weld electrodes. Use the following vertical shear loads: PD = 9 kips PL = 27 kips

Fig K.4-1. Connection geometry for Example K.4. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Column ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi

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From AISC Manual Tables 1-1 and 1-12, the geometric properties are as follows: W2162

tw = 0.400 in. d = 21.0 in. kdes = 1.12 in. HSS12122

t = 0.465 in. B = 12.0 in. From of ASCE/SEI 7, Chapter 2, the required strength is: ASD

LRFD Ru  1.2  9 kips   1.6  27 kips 

Ra  9 kips  27 kips  36.0 kips

 54.0 kips

Seat Angle and Weld Design Check web local yielding of the W2162 using AISC Manual Part 9. LRFD From AISC Manual Equation 9-46a and Table 9-4:

Ru  R1  kdes R2 54.0 kips  56.0 kips  20.0 kip/in.

ASD From AISC Manual Equation 9-46b and Table 9-4:

Ra  R1 /   kdes R2 /  36.0 kips  37.3 kips  13.3 kip/in.

lb min 

lb min 

which results in a negative quantity.

which results in a negative quantity.

Use lb min = kdes = 1.12 in.

Use lb min = kdes = 1.12 in.

Check web local crippling when lb/d M 0.2.

Check web local crippling when lb/d M 0.2.

From AISC Manual Equation 9-48a:

From AISC Manual Equation 9-48b: Ra  R3 /  R4 /  36.0 kips  47.8 kips  3.58 kip/in.

Ru  R3 R4 54.0 kips  71.7 kips  5.37 kip/in.

lb min 

lb min 

which results in a negative quantity.

which results in a negative quantity.

Check web local crippling when lb/d > 0.2.

Check web local crippling when lb/d > 0.2.

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LRFD From AISC Manual Equation 9-49a:

ASD From AISC Manual Equation 9-49b:

Ru  R5 R6 54.0 kips  64.2 kips  7.16 kip/in.

Ra  R5 /  R6 /  36.0 kips  42.8 kips  4.77 kip/in.

lb min 

lb min 

which results in a negative quantity.

which results in a negative quantity.

Note: Generally, the value of lb/d is not initially known and the larger value determined from the web local crippling equations in the preceding text can be used conservatively to determine the bearing length required for web local crippling. For this beam and end reaction, the beam web available strength exceeds the required strength (hence the negative bearing lengths) and the lower-bound bearing length controls (lb req = kdes = 1.12 in.). Thus, lb min = 1.12 in. Try an L84s seat with c-in. fillet welds. Outstanding Angle Leg Available Strength From AISC Manual Table 10-6 for an 8-in. angle length and lb req = 1.12 in.  18 in., the outstanding angle leg available strength is: LRFD Rn  81.0 kips  54.0 kips

ASD Rn  53.9 kips  36.0 kips o.k. 

o.k.

Available Weld Strength From AISC Manual Table 10-6, for an 8 in. x 4 in. angle and c-in. weld size, the available weld strength is: LRFD Rn  66.7 kips  54.0 kips

ASD Rn  44.5 kips  36.0 kips o.k. 

o.k.

Minimum HSS Wall Thickness to Match Weld Strength tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  5 

62 ksi  0.249 in.  0.465 in.

Because t of the HSS is greater than tmin for the c-in. weld, no reduction in the weld strength is required to account for the shear in the HSS. Connection to Beam and Top Angle (AISC Manual Part 10) Use a L444 top angle for stability. Use a x-in. fillet weld across the toe of the angle for attachment to the HSS. Attach both the seat and top angles to the beam flanges with two w-in.-diameter Group A bolts.

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EXAMPLE K.5

STIFFENED SEATED CONNECTION TO AN HSS COLUMN

Given:

Use AISC Manual Tables 10-8 and 10-15 to verify a stiffened seated connection for an ASTM A992 W2168 beam to an ASTM A500 Grade C HSS14142 column, as shown in Figure K.5-1. Use 70-ksi electrode welds to connect the stiffener, seat plate and top angle to the HSS. The angle and plate material are ASTM A36. Use the following vertical shear loads: PD = 20 kips PL = 60 kips

Fig K.5-1. Connection geometry for Example K.5. Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi

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Column ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi Angles and Plates ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-1 and 1-12, the geometric properties are as follows: W2168 tw = 0.430 in. d = 21.1 in. kdes = 1.19 in. HSS14142

t = 0.465 in. B = 14.0 in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2  20 kips   1.6  60 kips 

ASD Pa  20 kips  60 kips  80.0 kips

 120 kips

The available strength of connections to rectangular HSS with concentrated loads are determined based on the applicable limit states from Chapter J. Stiffener Width, W, Required for Web Local Crippling and Web Local Yielding The stiffener width is determined based on web local crippling and web local yielding of the beam, assuming a w-in. beam end setback in the calculations. Note that according to AISC Specification Section J10, the length of bearing, lb, cannot be less than the beam kdes. For web local crippling, assume lb/d > 0.2 and use constants R5 and R6 from AISC Manual Table 9-4. LRFD From AISC Manual Equation 9-49a and Table 9-4:

Ru  R5  setback  kdes  setback R6 120 kips  75.9 kips   w in.  1.19 in.  w in. 7.95 kip/in.  6.30 in.  1.94 in.

Wmin 

ASD From AISC Manual Equation 9-49b and Table 9-4:

Ra  R5 /   setback  kdes  setback R6 /  80.0 kips  50.6 kips   w in.  1.19 in.  w in. 5.30 kip/in.  6.30 in.  1.94 in.

Wmin 

For web local yielding, use constants R1 and R2 from AISC Manual Table 9-4.

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LRFD From AISC Manual Equation 9-46a and Table 9-4: Ru  R1  setback  kdes  setback R2 120 kips  64.0 kips   w in.  1.19 in.  w in. 21.5 kip/in.  3.35 in.  1.94 in.

Wmin 

ASD From AISC Manual Equation 9-46a and Table 9-4:

Ra  R1 /   setback  kdes  setback R2 /  80.0 kips  42.6 kips   w in.  1.19 in.  w in. 14.3 kip/in.  3.37 in.  1.94 in.

Wmin 

The minimum stiffener width, Wmin, for web local crippling controls. The stiffener width of 7 in. is adequate. Check the assumption that lb/d > 0.2. lb  7 in.  w in.  6.25 in.

lb 6.25 in.  d 21.1 in.  0.296  0.2, as assumed Weld Strength Requirements for the Seat Plate Check the stiffener length, l = 24 in., with c-in. fillet welds. Enter AISC Manual Table 10-8, using W = 7 in. as verified in the preceding text. LRFD Rn  293 kips  120 kips

ASD Rn  195 kips  80.0 kips 

o.k.

o.k.

From AISC Manual Part 10, Figure 10-10(b), the minimum length of the seat-plate-to-HSS weld on each side of the stiffener is 0.2l = 4.80 in. This establishes the minimum weld between the seat plate and stiffener. A 5-in.-long cin. weld on each side of the stiffener is adequate. Minimum HSS Wall Thickness to Match Weld Strength The minimum HSS wall thickness required to match the shear rupture strength of the base metal to that of the weld is: 3.09 D tmin  (Manual Eq. 9-2) Fu 

3.09  5 

62 ksi  0.249 in.  0.465 in.

Because t of the HSS is greater than tmin for the c-in. fillet weld, no reduction in the weld strength to account for shear in the HSS is required. Stiffener Plate Thickness From AISC Manual Part 10, Table 10-8 discussion, to develop the stiffener-to-seat-plate welds, the minimum stiffener thickness is:

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t p min  2 w  2  c in.  s in. Also, from AISC Manual Part 10, Table 10-8 discussion, for a stiffener with Fy = 36 ksi and a beam with Fy = 50 ksi, the minimum stiffener thickness is:  Fy beam  t p min    tw  Fy stiffener   50 ksi     0.430 in.  36 ksi   0.597 in.

The stiffener thickness of s in. is adequate. Determine the stiffener length using AISC Manual Table 10-15. The required HSS wall strength factor is:

 RuW   2    t req

LRFD 120 kips   7 in.

 0.465 in.

 RaW   2    t  req

2

 3,880 kip/in.

ASD 80.0 kips   7 in.

 0.465 in.2

 2,590 kip/in.

To satisfy the minimum, select a stiffener with l = 24 in. from AISC Manual Table 10-15. The HSS wall strength factor is: LRFD RuW t2

ASD

 3,910 kip/in.  3,880 kip/in. o.k.

RaW t2

 2, 600 kip/in.  2,590 kip/in. o.k.

Use PLs in.7 in. 2 ft 0 in. for the stiffener. HSS Width Check The minimum width is 0.4l + tp + 2(2.25t); however, because the specified weld length of 5 in. on each side of the stiffener is greater than 0.4l, the weld length will be used. The nominal wall thickness, tnom, is used, as would be used to calculate a workable flat dimension.

B  14.0 in.   2 welds  5.00 in.  s in.  2  2.252 in.  14.0 in.  12.9 in. o.k. Seat Plate Dimensions To accommodate two w-in.-diameter Group A bolts on a 52-in. gage connecting the beam flange to the seat plate, a minimum width of 8 in. is required. To accommodate the seat-plate-to-HSS weld, the required width is: 2  5.00 in.  s in.  10.6 in.

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Note: To allow room to start and stop welds, an 11.5 in. width is used. Use PLa in.7 in.0 ft-112 in. for the seat plate. Top Angle, Bolts and Welds (AISC Manual Part 10) The minimum weld size for the HSS thickness according to AISC Specification Table J2.4 is x in. The angle thickness should be z in. larger. Use L444 with x-in. fillet welds along the toes of the angle to the beam flange and HSS for stability. Alternatively, two w-in.-diameter Group A bolts may be used to connect the leg of the angle to the beam flange.

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EXAMPLE K.6

SINGLE-PLATE CONNECTION TO A RECTANGULAR HSS COLUMN

Given:

Use AISC Manual Table 10-10a to verify the design of a single-plate connection for an ASTM A992 W1835 beam framing into an ASTM A500 Grade C HSS66a column, as shown in Figure K.6-1. Use 70-ksi weld electrodes. The plate material is ASTM A36. Use the following vertical shear loads: PD = 6.5 kips PL = 19.5 kips

Fig K.6-1. Connection geometry for Example K.6. Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Column ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-1 and 1-12, the geometric properties are as follows:

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W1835 d = 17.7 in. tw = 0.300 in. T = 152 in. HSS66a

B = H = 6.00 in. t = 0.349 in. b/t = 14.2 From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  6.5 kips   1.6 19.5 kips 

ASD Ra  6.5 kips  19.5 kips  26.0 kips

 39.0 kips

Single-Plate Connection As discussed in AISC Manual Part 10, a single-plate connection may be used as long as the HSS wall is not classified as a slender element.

b E  1.40 t Fy 14.2  1.40

29, 000 ksi 50 ksi

14.2  33.7 Therefore, the HSS wall is not slender. The available strength of the face of the HSS for the limit state of punching shear is determined from AISC Manual Part 10 as follows: LRFD

  0.75 

Ru e 

Fu tl p 2

(Manual Eq. 10-7a)

5

 39.0 kips  3 in. 

0.75  62 ksi  0.349 in. 8.50 in.

117 kip-in.  235 kip-in.

5 o.k.

ASD

  2.00 

2

Ra e 

Fu tl p 2 5

 26.0 kips  3 in. 

(Manual Eq. 10-7b)

 62 ksi  0.349 in. 8.50 in.2 5  2.00 

78.0 kip-in.  156 kip-in.

o.k.

Try three rows of bolts and a c-in. plate thickness with 4-in. fillet welds. From AISC Manual Table 10-9, either the plate or the beam web must satisfy: d  z in. 2 w in. c in.  + z in. 2 c in.  0.438 in. o.k.

t

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Obtain the available single-plate connection strength from AISC Manual Table 10-10a: ASD Rn  29.4 kips  26.0 kips o.k. 

LRFD Rn  44.2 kips  39.0 kips

o.k.

Use a PLc in.42 in. 0 ft 82 in. HSS Shear Rupture at Welds The minimum HSS wall thickness required to match the shear rupture strength of the HSS wall to that of the weld is: tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  4 

62 ksi  0.199 in.  t  0.349 in.

o.k.

Available Beam Web Strength The available beam web strength is the lesser of the limit states of block shear rupture, shear yielding, shear rupture, and the sum of the effective strengths of the individual fasteners. The beam is not coped, so the only applicable limit state is the effective strength of the individual fasteners. The effective strength of an individual fastener is the lesser of the fastener shear strength, the bearing strength at the bolt hole and the tearout strength at the bolt hole. For the limit state of fastener shear strength, with Ab = 0.442 in.2 from AISC Manual Table 7-1 for a w-in. bolt.: rn  Fnv Ab



  54 ksi  0.442 in.

2

(from Spec. Eq. J3-1)



 23.9 kips/bolt

where Fnv is the nominal shear strength of a Group A bolt in a bearing-type connection when threads are not excluded from the shear plane as found in AISC Specification Table J3.2. Assume that deformation at the bolt hole at service load is a design consideration. For the limit state of bearing: rn  2.4dtFu

(from Spec. Eq. J3-6a)

 2.4  w in. 0.300 in. 65 ksi   35.1 kips/bolt For the limit state of tearout: rn  1.2lc tFu

(from Spec. Eq. J3-6c)

 1.2  3 in.  m in. 0.300 in. 65 ksi   51.2 kips/bolt where lc is the clear distance, in the direction of the force, between the edges of the bolt holes.

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Fastener shear strength is the governing limit state for all bolts at the beam web. Fastener shear strength is one of the limit states included in the available strengths given in Table 10-10a and used in the preceding calculations. Thus, the effective strength of the fasteners is adequate.

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EXAMPLE K.7

THROUGH-PLATE CONNECTION TO A RECTANGULAR HSS COLUMN

Given:

Use AISC Manual Table 10-10a to verify a through-plate connection between an ASTM A992 W1835 beam and an ASTM A500 Grade C HSS648 with the connection to one of the 6 in. faces, as shown in Figure K.7-1. A thin-walled column is used to illustrate the design of a through-plate connection. Use 70-ksi weld electrodes. The plate is ASTM A36 material. Use the following vertical shear loads: PD = 3.3 kips PL = 9.9 kips

Fig K.7-1. Connection geometry for Example K.7. Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Column ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-1 and 1-11, the geometric properties are as follows:

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W1835

d = 17.7 in. tw = 0.300 in. T = 152 in. HSS648 B = 4.00 in. H = 6.00 in. t = 0.116 in. h/t = 48.7 b/t = 31.5

HSS wall slenderness From AISC Manual Part 10, the limiting width-to-thickness for a nonslender HSS wall is:

1.40

E 29, 000 ksi  1.40 Fy 50 ksi  33.7

Because h/t = 48.7 > 33.7, the HSS648 is slender and a through-plate connection should be used instead of a single-plate connection. Through-plate connections are typically very expensive. When a single-plate connection is not adequate, another type of connection, such as a double-angle connection may be preferable to a through-plate connection. AISC Specification Chapter K does not contain provisions for the design of through-plate shear connections. The following procedure treats the connection of the through-plate to the beam as a single-plate connection. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  3.3 kips   1.6  9.9 kips 

 19.8 kips

ASD Ra  3.3 kips  9.9 kips  13.2 kips

Portion of the Through-Plate Connection that Resembles a Single-Plate Try three rows of bolts (l = 82 in.) and a 4-in. plate thickness with x-in. fillet welds. T 152 in.  2 2  7.75 in.  l  82 in. o.k.

Note: From AISC Manual Table 10-9, the larger of the plate thickness or the beam web thickness must satisfy: d  z in. 2 w in. 4 in.   z in. 2 4 in.  0.438 in. o.k. t

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Obtain the available single-plate connection strength from AISC Manual Table 10-10a: LRFD

ASD Rn  25.6 kips  13.2 kips 

Rn  38.3 kips  19.8 kips o.k.

o.k.

Required Weld Strength The available strength for the welds in this connection is checked at the location of the maximum reaction, which is along the weld line closest to the bolt line. The reaction at this weld line is determined by taking a moment about the weld line farthest from the bolt line. a = 3 in. (distance from bolt line to nearest weld line)

V fu  

Ru  B  a 

LRFD V fa 

B 19.8 kips  4.00 in.  3 in.



4.00 in.

Ra  B  a 

ASD

B 13.2 kips  4.00 in.  3 in. 4.00 in.

 23.1 kips

 34.7 kips

Available Weld Strength The minimum required weld size is determined using AISC Manual Part 8. LRFD Dreq

V fu  1.392l 

ASD (from Manual Eq. 8-2a)

34.7 kips 1.392 kip/in.  8.50 in. 2 

 1.47 sixteenths  3 sixteenths

Dreq

V fa  0.928l 

o.k.

(from Manual Eq. 8-2b)

23.1 kips 0.928 kip/in.  8.50 in. 2 

 1.46 sixteenths  3 sixteenths

o.k.

HSS Shear Yielding and Rupture Strength The available shear yielding strength of the HSS is determined from AISC Specification Section J4.2.   1.00

LRFD

Rn  0.60 Fy Agv

  1.50 (from Spec. Eq. J4-3)

 1.00  0.60  50 ksi  0.116 in. 8.50 in. 2   59.2 kips  34.7 kips o.k.

ASD

Rn 0.60 Fy Agv (from Spec. Eq. J4-3)     0.60  50 ksi  0.116 in.8.50 in. 2   1.50  39.4 kips  23.1 kips o.k.

The available shear rupture strength of the HSS is determined from AISC Specification Section J4.2.

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LRFD

  0.75 Rn  0.60 Fu Anv

  2.00 (from Spec. Eq. J4-4)

 0.75  0.60  62 ksi  0.116 in. 8.50 in. 2   55.0 kips  34.7 kips

o.k.

ASD

Rn 0.60 Fu Anv (from Spec. Eq. J4-4)     0.60  62 ksi  0.116 in.8.50 in. 2   2.00  36.7 kips  23.1 kips o.k.

Available Beam Web Strength The available beam web strength is the lesser of the limit states of block shear rupture, shear yielding, shear rupture, and the sum of the effective strengths of the individual fasteners. The beam is not coped, so the only applicable limit state is the effective strength of the individual fasteners. The effective strength of an individual fastener is the lesser of the fastener shear strength, the bearing strength at the bolt hole and the tearout strength at the bolt hole. For the limit state of fastener shear strength, with Ab = 0.442 in.2 from AISC Manual Table 7-1 for a w-in. bolt: rn  Fnv Ab



  54 ksi  0.442 in.

2

(from Spec. Eq. J3-1)



 23.9 kips/bolt

where Fnv is the nominal shear strength of a Group A bolt in a bearing-type connection when threads are not excluded from the shear planes as found in AISC Specification Table J3.2. Assume that deformation at the bolt hole at service load is a design consideration. For the limit state of bearing: rn  2.4dtFu

(from Spec. Eq. J3-6a)

 2.4  w in. 0.300 in. 65 ksi   35.1 kips/bolt For the limit state of tearout: rn  1.2lc tFu

(from Spec. Eq. J3-6c)

 1.2  3 in.  m in. 0.300 in. 65 ksi   51.2 kips/bolt where lc is the clear distance, in the direction of the force, between the edges of the bolt holes. Fastener shear strength is the governing limit state for all bolts at the beam web. Fastener shear strength is one of the limit states included in the available strengths shown in Table 10-10a as used in the preceding calculations. Thus, the effective strength of the fasteners is adequate.

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EXAMPLE K.8 ROUND HSS

LONGITUDINAL PLATE LOADED PERPENDICULAR TO THE HSS AXIS ON A

Given:

Verify the local strength of the ASTM A500 Grade C HSS6.0000.375 tension chord subject to transverse loads, PD = 4 kips and PL = 12 kips, applied through an ASTM A36 plate, as shown in Figure K.8-1.

Fig K.8-1. Loading and geometry for Example K.8. Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Chord ASTM A500 Grade C Fy = 46 ksi Fu = 62 ksi Plate ASTM A36 Fyp = 36 ksi Fu = 58 ksi From AISC Manual Table 1-13, the geometric properties are as follows: HSS6.0000.375

D = 6.00 in. t = 0.349 in. D/t = 17.2 Limits of Applicability of AISC Specification Section K2.2, Table K2.1A AISC Specification Table K2.1A provides the limits of applicability for plate-to-round connections. The applicable limits for this example are: HSS wall slenderness: D t  50 for T-connections 17.2  50 o.k.

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Material strength: Fy  52 ksi 46 ksi  52 ksi

o.k.

Ductility: Fy  0.8 Fu 46 ksi  0.8 62 ksi 0.741  0.8 o.k. End distance: B D  lend  D  1.25  b  2    4 in. 6.00 in.    6.00 in. 1.25   2    7.38 in. Thus, the edge of the plate must be located a minimum of 7.38 in. from the end of the HSS. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2  4 kips   1.6 12 kips 

ASD

Pa  4 kips  12 kips  16.0 kips

 24.0 kips HSS Plastification Limit State

The limit state of HSS plastification applies and is determined from AISC Specification Table K2.1. l   Rn sin   5.5Fy t 2  1  0.25 b  Q f D 

(Spec. Eq. K2-2a)

From the AISC Specification Table K2.1 Functions listed at the bottom of the table, for an HSS connecting surface in tension, Qf = 1.0. 2  4 in.   5.5  46 ksi  0.349 in. 1  0.25    1.0   6.00 in.    Rn  sin 90  36.0 kips

The available strength is:

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  0.90 

LRFD

 Rn  0.90  36.0 kips 

 32.4 kips  24.0 kips o.k.

  1.67 

ASD

 

Rn 36.0 kips   1.67  21.6 kips  16.0 kips o.k.

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EXAMPLE K.9

RECTANGULAR HSS COLUMN BASE PLATE

Given: An ASTM A500 Grade C HSS662 column is supporting loads of 40 kips of dead load and 120 kips of live load. The column is supported by a 7 ft 6 in.  7 ft 6 in. concrete spread footing with f c = 3,000 psi. Verify the ASTM A36 base plate size shown in Figure K.9-1 for this column.

Fig K.9-1. Base plate geometry for Example K.9. Solution: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Column ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi Base Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-12, the geometric properties are as follows: HSS662

B = H = 6.00 in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2  40 kips   1.6 120 kips 

 240 kips

ASD

Pa  40 kips  120 kips  160 kips

Note: The procedure illustrated here is similar to that presented in AISC Design Guide 1, Base Plate and Anchor Rod Design (Fisher and Kloiber, 2006), and AISC Manual Part 14.

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Try a base plate which extends 32 in. from each face of the HSS column, or 13 in.  13 in. Available Strength for the Limit State of Concrete Crushing On less than the full area of a concrete support:

Pp  0.85 fcA1 A2 A1  1.7fcA1

(Spec. Eq. J8-2)

A1  BN  13 in.13 in.  169 in.2 A2   7.5 ft 12 in./ft  

2

 8,100 in.2



Pp  0.85  3 ksi  169 in.2



8,100 in.2 2

169 in.



 1.7  3 ksi  169 in.2



 2,980 kips  862 kips Use Pp = 862 kips. Note: The limit on the right side of AISC Specification Equation J8-2 will control when A2/A1 exceeds 4.0. LRFD From AISC Specification Section J8: c  0.65

ASD From AISC Specification Section J8:  c  2.31

c Pp  0.65  862 kips 

Pp 862 kips  c 2.31  373 kips  160 kips

 560 kips  240 kips o.k.

o.k.

Pressure under Bearing Plate and Required Thickness For a rectangular HSS, the distance m or n is determined using 0.95 times the depth and width of the HSS. mn  

(from Manual Eq. 14-2)

N  0.95  B or H  2 13 in.  0.95  6.00 in. 2

 3.65 in.

Note: As discussed in AISC Design Guide 1, the n cantilever distance is not used for HSS and pipe. The critical bending moment is the cantilever moment outside the HSS perimeter. Therefore, m = n = l.

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LRFD f pu  

ASD

Pu A1 240 kips

f pa  

169 in.2  1.42 ksi

Z

169 in.2  0.947 ksi

f pu l 2

Mu 

Pa A1 160 kips

Ma 

2

t p2

Z

4

f pa l 2 2 t p2 4

b = 1.67

b = 0.90 Mn = Mp = FyZ

(from Spec. Eq. F11-1)

Mn = Mp = FyZ

(from Spec. Eq. F11-1)

Note: the upper limit of 1.6FySx will not govern for a rectangular plate.

Note: the upper limit of 1.6FySx will not govern for a rectangular plate.

Equating:

Equating:

Mu = bMn and solving for tp gives:

Ma = Mn/b and solving for tp gives:

t p ( req )  

2 f pu l 2 b Fy

t p ( req ) 

2 1.42 ksi  3.65 in.

2



0.90  36 ksi 

 1.08 in.

2

 36 ksi  / 1.67

Or use AISC Manual Equation 14-7b:

2 Pu 0.90 Fy BN

  3.65 in.

2  0.947 ksi  3.65 in.

 1.08 in.

Or use AISC Manual Equation 14-7a: tmin  l

2 f pa l 2 Fy / b

tmin  l 2  240 kips 

0.90  36 ksi 13 in.13 in 

1.08 in.

1.67  2 Pa  Fy BN

  3.65 in.

1.67  2 160 kips 

 36 ksi 13 in.13 in.

1.08 in.

Therefore, the PL14 in. 13 in. 1 ft 1 in. is adequate.

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EXAMPLE K.10 RECTANGULAR HSS STRUT END PLATE Given: Determine the weld leg size, end-plate thickness, and the bolt size required to resist forces of 16 kips from dead load and 50 kips from live load on an ASTM A500 Grade C section, as shown in Figure K.10-1. The end plate is ASTM A36. Use 70-ksi weld electrodes.

Fig K.10-1. Loading and geometry for Example K.10.

Solution: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Strut ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi End Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-12, the geometric properties are as follows: HSS444

t = 0.233 in. A = 3.37 in.2 From ASCE/SEI 7, Chapter 2, the required tensile strength is: LRFD Pu  1.2 16 kips   1.6  50 kips 

 99.2 kips

ASD

Pa  16 kips  50 kips  66.0 kips

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Preliminary Size of the (4) Group A Bolts ASD

LRFD Pu n 99.2 kips  4  24.8 kips

Pa n 66.0 kips  4  16.5 kips

rut 

rat 

Using AISC Manual Table 7-2, try w-in.-diameter Group A bolts.

Using AISC Manual Table 7-2, try w-in.-diameter Group A bolts.

rn  29.8 kips

rn  19.9 kips 

End-Plate Thickness with Consideration of Prying Action (AISC Manual Part 9) d  a   a  b 2 

db      1.25b   2    w in. w in.  12 in.   1.25 12 in.  2 2  1.88 in.  2.25 in.  1.88 in.

b  b 

db 2

 12 in. 

(Manual Eq. 9-23)

(Manual Eq. 9-18) w in. 2

 1.13 in. b a 1.13  1.88  0.601

(Manual Eq. 9-22)



d   m in.

The tributary length per bolt (Packer et al., 2010),

full plate width number of bolts per side 10.0 in.  1  10.0 in.

p

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d p m in.  1 10.0 in.  0.919

  1

(Manual Eq. 9-20)

LRFD 1  rn  (from Manual Eq. 9-21)    1   rut  1  29.8 kips    1  0.601  24.8 kips   0.335 Because  < 1, from AISC Manual Part 9:

1        1.0  1   1  0.335      1.0 0.919  1  0.335   0.548

ASD 1r /    n  1   rat 

(from Manual Eq. 9-21)

1  19.9 kips   1  0.601  16.5 kips   0.343 

Because  < 1, from AISC Manual Part 9:  

1      1.0  1  

1  0.343     1.0 0.919  1  0.343   0.568 

Use Equation 9-19 for tmin in Chapter 9 of the AISC Manual, except that Fu is replaced by Fy per the recommendation of Willibald, Packer and Puthli (2003) and Packer et al. (2010). ASD

LRFD tmin  

4rut b pFy 1   

(from Manual Eq. 9-19a)

4  24.8 kips 1.13 in.

0.90 10.0 in. 36 ksi  1  0.919  0.548  

tmin  

4rat b pFy (1  )

(from Manual Eq. 9-19b)

1.67  4 16.5 kips 1.13 in.

10.0 in. 36 ksi  1  0.919  0.568 

 0.477 in.

 0.480 in.

Use a 2-in.-thick end plate, t1 > 0.480 in., further bolt check for prying not required.

Use a 2-in.-thick end plate, t1 > 0.477 in., further bolt check for prying not required.

Use (4) w-in.-diameter Group A bolts.

Use (4) w-in.-diameter Group A bolts.

Required Weld Size Rn  Fnw Awe

(Spec. Eq. J2-4)



Fnw  0.60 FEXX 1.0  0.50sin1.5 





(Spec. Eq. J2-5)

 0.60  70 ksi  1.0  0.50sin1.5 90



 63.0 ksi

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 2  D  Awe      l  2   16  where D is the weld size in sixteenths of an inch (i.e., D is an integer). l  4  4.00 in.  16.0 in.

Note: This weld length is approximate. A more accurate length could be determined by taking into account the curved corners of the HSS. From AISC Specification Table J2.5: LRFD

  0.75 

 Rn  Fnw Awe  2  D    0.75  63.0 ksi      16.0 in.  2   16 

Rn  Fnw Awe       



Setting Rn  Pu and solving for D, D

D = 3 (i.e., a x in. weld)

D

 2  D     16.0 in.  2   16 

 63.0 ksi  

Setting

 99.2 kips 16 

 2 0.75  63.0 ksi    16.0 in.  2   2.97

ASD

  2.00 

2.00

Rn  Pa and solving for D,  2.00  66.0 kips 16   2  16.0 in.  2 

 63.0 ksi    2.96

D = 3 (i.e., a x in. weld) Minimum Weld Size Requirements For t = 4 in., the minimum weld size = 8 in. from AISC Specification Table J2.4. Summary: Use a x-in. weld with 2-in.-thick end plates and (4) w-in.-diameter Group A bolts.

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CHAPTER K DESIGN EXAMPLE REFERENCES Fisher, J.M. and Kloiber, L.A. (2006), Base Plate and Anchor Rod Design, Design Guide 1, 2nd Ed., AISC, Chicago, IL Packer, J.A., Sherman, D. and Lecce, M. (2010), Hollow Structural Section Connections, Design Guide 24, AISC, Chicago, IL. Willibald, S., Packer, J.A. and Puthli, R.S. (2003), “Design Recommendations for Bolted Rectangular HSS Flange Plate Connections in Axial Tension,” Engineering Journal, AISC, Vol. 40, No. 1, pp. 15–24.

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APPENDIX 6 MEMBER STABILITY BRACING This Appendix addresses the minimum strength and stiffness necessary to provide a braced point in a column, beam or beam-column. The governing limit states for column and beam design may include flexural, torsional and flexural-torsional buckling for columns and lateral-torsional buckling for beams. In the absence of other intermediate bracing, column unbraced lengths are defined between points of obviously adequate lateral restraint, such as floor and roof diaphragms that are part of the building’s lateral force-resisting systems. Similarly, beams are often braced against lateral-torsional buckling by relatively strong and stiff bracing elements such as a continuously connected floor slab or roof diaphragm. However, at times, unbraced lengths are bounded by elements that may or may not possess adequate strength and stiffness to provide sufficient bracing. AISC Specification Appendix 6 provides equations for determining the required strength and stiffness of braces that have not been included in the second-order analysis of the structural system. It is not intended that the provisions of Appendix 6 apply to bracing that is part of the lateral force-resisting system. Guidance for applying these provisions to stabilize trusses is provided in AISC Specification Appendix 6 commentary. Background for the provisions can be found in references cited in the Commentary including “Fundamentals of Beam Bracing” (Yura, 2001) and the Guide to Stability Design Criteria for Metal Structures (Ziemian, 2010). AISC Manual Part 2 also provides information on member stability bracing. 6.1

GENERAL PROVISIONS

Lateral column and beam bracing may be either panel or point while torsional beam bracing may be point or continuous. The User Note in AISC Specification Appendix 6, Section 6.1 states “A panel brace (formerly referred to as a relative brace) controls the angular deviation of a segment of the braced member between braced points (that is, the lateral displacement of one end of the segment relative to the other). A point brace (formerly referred to as a nodal brace) controls the movement at the braced point without direct interaction with adjacent braced points. A continuous bracing system consists of bracing that is attached along the entire member length.” Panel and point bracing systems are discussed further in AISC Specification Commentary Appendix 6, Section 6.1. Examples of each bracing type are shown in AISC Specification Commentary Figure C-A-6.1. In lieu of the requirements of Appendix 6, Sections 6.2, 6.3 and 6.4, alternative provisions are given in Sections 6.1(a), 6.1(b) and 6.1(c). 6.2

COLUMN BRACING

The requirements in this section apply to bracing associated with the limit state of flexural buckling. For columns that could experience torsional or flexural-torsional buckling, as addressed in AISC Specification Section E4, the designer must ensure that sufficient bracing to resist the torsional component of buckling is provided. See Helwig and Yura (1999). Column braces may be panel or point. The type of bracing must be determined before the requirements for strength and stiffness can be determined. The requirements are derived for an infinite number of braces along the column and are thus conservative for most columns as explained in the Commentary. Provision is made in this section for reducing the required brace stiffness for point bracing when the column required strength is less than the available strength of the member. The Commentary also provides an approach to reduce the requirements when a finite number of point braces are provided. 6.3

BEAM BRACING

The requirements in this section apply to bracing of doubly and singly symmetric I-shaped members subject to flexure within a plane of symmetry and zero net axial force. Bracing to resist lateral-torsional buckling may be

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accomplished by a lateral brace, a torsional brace, or a combination of the two to prevent twist of the section. Lateral bracing should normally be connected near the compression flange. The exception is for the free ends of cantilevers and near inflection points of braced beams subject to double curvature bending. Torsional bracing may be connected anywhere on the cross section in a manner to prevent twist of the section. According to AISC Specification Section F1(b), the design of members for flexure is based on the assumption that points of support are restrained against rotation about their longitudinal axis. The bracing requirements in Appendix 6 are for intermediate braces in addition to those at the support. In members subject to double curvature, inflection points are not to be considered as braced points unless bracing is provided at that location. In addition, the bracing nearest the inflection point must be attached to prevent twist, either as a torsional brace or as lateral braces attached to both flanges as described in AISC Specification Appendix 6, Section 6.3.1(b). 6.3.1

Lateral Bracing

As with column bracing, beam bracing may be panel or point. In addition, it is permissible to provide torsional bracing. This section provides requirements for determining the required lateral brace strength and stiffness for panel and point braces. For point braces, provision is made in this section to reduce the required brace stiffness when the actual unbraced length is less than the maximum unbraced length for the required flexural strength. 6.3.2

Torsional Bracing

This section provides requirements for determining the required bracing flexural strength and stiffness for point and continuous torsional bracing. Torsional bracing can be connected to the section at any cross-section location. However, if the beam has inadequate distortional (out-of-plane) bending stiffness, torsional bracing will be ineffective. Web stiffeners can be provided when necessary, to increase the web distortional stiffness for point torsional braces. As is the case for columns and for lateral beam point braces, it is possible to reduce the required brace stiffness when the required strength of the member is less than the available strength for the provided location of bracing. Provisions for continuous torsional bracing are also provided. A slab connected to the top flange of a beam in double curvature may provide sufficient continuous torsional bracing as discussed in the Commentary. For this condition there is no unbraced length between braces so the unbraced length used in the strength and stiffness equations is the maximum unbraced length permitted to provide the required strength in the beam. In addition, for continuous torsional bracing, stiffeners are not permitted to be used to increase web distortional stiffness. 6.4

BEAM-COLUMN BRACING

For bracing of beam-columns, the required strength and stiffness are to be determined for the column and beam independently as specified in AISC Specification Appendix 6, Sections 6.2 and 6.3. These values are then to be combined, depending on the type of bracing provided.

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EXAMPLE A-6.1

POINT STABILITY BRACING OF A W-SHAPE COLUMN

Given: Determine the required strength and the stiffness for intermediate point braces, such that the unbraced length for the column can be taken as 12 ft. The column is an ASTM A992 W1272 with loading and geometry as shown in Figure A-6.1-1. The column is braced laterally and torsionally at its ends with intermediate lateral braces for the xand y-axis provided at the one-third points as shown. Thus, the unbraced length for the limit state of flexuraltorsional buckling is 36 ft and the unbraced length for flexural buckling is 12 ft. The column has sufficient strength to support the applied loads with this bracing.

Fig. A-6.1-1. Column bracing geometry for Example A-6.1. Solution: From AISC Manual Table 2-4, the material properties are as follows: Column ASTM A992 Fy = 50 ksi Fu = 65 ksi Required Compressive Strength of Column From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2 105 kips   1.6  315 kips 

ASD Pa  105 kips  315 kips

 420 kips

 630 kips

Available Compressive Strength of Column From AISC Manual Table 4-1a at Lcy = 12 ft, the available strength of the W1272 is:

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LRFD c Pn  806 kips  630 kips

ASD Pn  536 kips  420 kips o.k. c

o.k.

Required Point Brace Strength From AISC Specification Appendix 6, Section 6.2.2, the required point brace strength is: LRFD

ASD

Pr  Pu

Pr  Pa

 630 kips

 420 kips

Pbr  0.01Pr

(Spec. Eq. A-6-3)

Pbr  0.01Pr

 0.01 630 kips 

 0.01 420 kips 

 6.30 kips

 4.20 kips

(Spec. Eq. A-6-3)

Required Point Brace Stiffness From AISC Specification Appendix 6, Section 6.2.2, the required point brace stiffness, with an unbraced length adjacent to the point brace Lbr = 12 ft, is:   0.75

LRFD

  2.00

Pr  Pa

Pr  Pu

 420 kips

 630 kips br 



1  8 Pr      Lbr 

ASD

(Spec. Eq. A-6-4a)

 8P  br    r   Lbr 

(Spec. Eq. A-6-4b)

 8  420 kips    2.00    12 ft 12 in./ft    46.7 kip/in.

1  8  630 kips     0.75  12 ft 12 in./ft  

 46.7 kip/in.

Determine the maximum permitted unbraced length for the required strength. Interpolating between values, from AISC Manual Table 4-1a: LRFD Lcy = 18.9 ft for Pu = 632 kips

ASD Lcy = 18.9 ft for Pa = 421 kips

Calculate the required point brace stiffness for this increased unbraced length It is permissible to design the braces to provide the lower stiffness determined using the maximum unbraced length permitted to carry the required strength according to AISC Specification Appendix 6, Section 6.2.2.

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  0.75

LRFD

  2.00

Pr  Pa

Pr  Pu

 420 kips

 630 kips br  

1  8 Pr    Lbr  1  8  630 kips     0.75  18.9 ft 12 in./ft  

 29.6 kip/in.

ASD

(Spec. Eq. A-6-4a)

 8P  br    r   Lbr   8  420 kips    2.00    18.9 ft 12 in./ft    29.6 kip/in.

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EXAMPLE A-6.2 POINT STABILITY BRACING OF A WT-SHAPE COLUMN Given:

Determine the strength and stiffness requirements for the point braces and select a W-shape brace based on x-axis flexural buckling of the ASTM A992 WT734 column with loading and geometry as shown in Figure A-6.2-1. The unbraced length for this column is 7.5 ft. Bracing about the y-axis is provided by the axial resistance of a W-shape connected to the flange of the WT, while bracing about the x-axis is provided by the flexural resistance of the same W-shape loaded at the midpoint of a 12-ft-long simple span beam. Assume that the axial strength and stiffness of the W-shape are adequate to brace the y-axis of the WT. Also, assume the column is braced laterally and torsionally at its ends and is torsionally braced at one-quarter points by the W-shape braces.

(a) Plan

(b) Elevation

Fig. A-6.2-1. Column bracing geometry for Example A-6.2. Solution:

From AISC Manual Table 2-4, the material properties of the column and brace are as follows: ASTM A992 Fy = 50 ksi Fu = 65 ksi Required Compressive Strength of Column From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2  25 kips   1.6  75 kips 

ASD Pa  25 kips  75 kips  100 kips

 150 kips

Available Compressive Strength of Column Interpolating between values, from AISC Manual Table 4-7, the available axial compressive strength of the WT734 with Lcx = 7.5 ft is:

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LRFD c Pn  357 kips  150 kips

ASD Pn  238 kips  100 kips c

o.k.

o.k.

Required Point Brace Size From AISC Specification Appendix 6, Section 6.2.2, the required point brace strength is: LRFD

ASD

Pr  Pu

Pr  Pa

 150 kips

 100 kips

Pbr  0.01Pr

(Spec. Eq. A-6-3)

Pbr  0.01Pr

 0.01150 kips 

 0.01100 kips 

 1.50 kips

 1.00 kips

(Spec. Eq. A-6-3)

From AISC Specification Appendix 6, Section 6.2.2, the required point brace stiffness is:   0.75

LRFD

  2.00

Pr  Pa

Pr  Pu

 100 kips

 150 kips br 



1  8 Pr      Lbr 

ASD

(Spec. Eq. A-6-4a)

 8P  br    r   Lbr 

(Spec. Eq. A-6-4b)

 8 100 kips    2.00     7.50 ft 12 in./ft  

1  8 150 kips     0.75   7.50 ft 12 in./ft  

 17.8 kip/in.

 17.8 kip/in.

The brace is a simple-span beam loaded at its midspan. Thus, its flexural stiffness can be derived from Case 7 of AISC Manual Table 3-23 to be 48EI/L3, which must be greater than the required point brace stiffness, br. Also, the flexural strength of the beam, bMp, for a compact laterally supported beam, must be greater than the moment resulting from the required brace strength over the beam’s simple span, Mbr = PbrL/4. Based on brace stiffness, the minimum required moment of inertia of the beam is:  L3 I br  br 48 E 

17.8 kip/in.12.0 ft 3 12 in./ft 3 48  29, 000 ksi 

 38.2 in.4

Based on moment strength for a compact laterally supported beam, the minimum required plastic section modulus is:

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LRFD Z req

ASD

M  br Fy 

Z req

1.50 kips 12.0 ft 12 in./ft  0.90(50 ksi)  4 

M br  Fy 

 1.20 in.3

1.67 1.00 kip 12.0 ft 12 in./ft 

 50 ksi  4 

 1.20 in.3

From AISC Manual Table 3-2, select a W813 member with Zx = 11.4 in.3 and Ix = 39.6 in.4 Note that because the live-to-dead load ratio is 3, the LRFD and ASD results are identical. The required stiffness can be reduced if the maximum permitted unbraced length is used as described in AISC Specification Appendix 6, Section 6.2, and also if the actual number of braces are considered, as discussed in the Commentary. The following demonstrates how this affects the design. Interpolating between values in AISC Manual Table 4-7, the maximum permitted unbraced length of the WT734 for the required strength is as follows: LRFD Lcx = 18.6 ft for Pu = 150 kips

ASD Lcx = 18.6 ft for Pa = 100 kips

From AISC Specification Commentary Appendix 6, Section 6.2, determine the reduction factor for three intermediate braces: 2n  1 2(3)  1  2n 2(3)  0.833

Determine the required point brace stiffness for the increased unbraced length and number of braces: LRFD

  0.75

  2.00

Pr  Pa

Pr  Pu

 100 kips

 150 kips

 1  8P br  0.833   r    Lbr

ASD

  

(Spec. Eq. A-6-4a)

 1  8(150 kips)    0.833     0.75  18.6 ft 12 in./ft     5.97 kip/in.

  8P   br  0.833   r     Lbr  

(Spec. Eq. A-6-4b)

  8(100 kips)    0.833 2.00    18.6 ft 12 in./ft      5.97 kip/in.

Determine the required brace size based on this new stiffness requirement. Based on brace stiffness, the minimum required moment of inertia of the beam is:

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I br  

br L3 48 E

 5.97 kip/in.12.0 ft 3 12 in./ft 3 48  29, 000 ksi 

 12.8 in.4

Based on the unchanged flexural strength for a compact laterally supported beam, the minimum required plastic section modulus, Zx, was determined previously to be 1.20 in.3 From AISC Manual Table 1-1, select a W68.5 noncompact member with Zx = 5.73 in.3 and Ix = 14.9 in.4

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EXAMPLE A-6.3

POINT STABILITY BRACING OF A BEAMCASE I

Given:

A walkway in an industrial facility has a span of 28 ft as shown in Figure A-6.3.1. The walkway has a deck of grating which is not sufficient to brace the beams. The ASTM A992 W1222 beams along walkway edges are braced against twist at the ends as required by AISC Specification Section F1(b) and are connected by an L334 strut at midspan. The two diagonal ASTM A36 L55c braces are connected to the top flange of the beams at the supports and at the strut at the middle. The strut and the brace connections are welded; therefore, bolt slippage does not need to be accounted for in the stiffness calculation. The dead load on each beam is 0.05 kip/ft and the live load is 0.125 kip/ft. Determine if the diagonal braces are strong enough and stiff enough to brace this walkway.

Fig. A-6.3-1. Plan view for Example A-6.3. Solution:

Because the diagonal braces are connected directly to an unyielding support that is independent of the midspan brace point, they are designed as point braces. The strut will be assumed to be sufficiently strong and stiff to force the two beams to buckle together. From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Diagonal braces ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-1 and 1-7, the geometric properties are as follows: Beam W1222 ho = 11.9 in. Diagonal braces L55c A = 3.07 in.2

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Required Flexure Strength of Beam From ASCE/SEI 7, Chapter 2, the required strength is: LRFD wu  1.2  0.05 kip/ft   1.6  0.125 kip/ft 

ASD wa  0.05 kip/ft  0.125 kip/ft  0.175 kip/ft

 0.260 kip/ft

Determine the required flexural strength for a uniformly loaded simply supported beam using AISC Manual Table 3-23, Case 1. LRFD Mu  

ASD

2

wu L 8

Ma 

 0.260 kip/ft  28 ft 2



8  25.5 kip-ft

2

wa L 8

 0.175 kip/ft  28 ft 2 8

 17.2 kip-ft

It can be shown that the W1222 beams are adequate with the unbraced length of 14 ft. Both beams need bracing in the same direction simultaneously. Required Brace Strength and Stiffness From AISC Specification Appendix 6, Section 6.3, determine the required point brace strength for each beam as follows, with Cd = 1.0 for bending in single curvature. LRFD

ASD

Mr  Mu

Mr  Ma

 25.5 kip-ft

 17.2 kip-ft

M C  Pbr  0.02  r d   ho 

(Spec. Eq. A-6-7)

  25.5 kip-ft 12 in. / ft 1.0    0.02   11.9 in.    0.514 kip

M C  Pbr  0.02  r d  (Spec. Eq. A-6-7)  ho   17.2 kip-ft 12 in. / ft 1.0    0.02   11.9 in.    0.347 kip

Because there are two beams to be braced, the total required brace strength is: Pbr  2  0.514 kip 

LRFD

Pbr  2  0.347 kip 

 1.03 kips

ASD

 0.694 kip

There are two beams to brace and two braces to share the load. The worst case for design of the braces will be when they are in compression. By geometry, the diagonal bracing length is

L

14 ft 2   5 ft 2

 14.9 ft

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The required brace strength is:

 5 ft  Pbr cos   Pbr    14.9 ft   1.03 kips

LRFD

ASD  5 ft  Pbr cos   Pbr    14.9 ft   0.694 kip

Because there are two braces, the required brace strength is:

Because there are two braces, the required brace strength is:

1.03 kips 2  5 ft 14.9 ft 

Pbr 

Pbr 

 1.53 kips

0.694 kip 2  5 ft 14.9 ft 

 1.03 kips

The required point brace stiffness, with Cd = 1.0 for bending in single curvature, is determined as follows: LRFD

  0.75

  2.00

Mr  Ma

Mr  Mu

 17.2 kip-ft

 25.5 kip-ft 1  10 M r Cd      Lbr ho 

br 



ASD

(Spec. Eq. A-6-8a)

 10 M r Cd  br      Lbr ho 

(Spec. Eq. A-6-8b)

10 17.2 kip-ft 12 in./ft 1.0    2.00    14 ft 12 in./ft 11.9 in. 

1 10  25.5 kip-ft 12 in./ft 1.0     0.75  14 ft 12 in./ft 11.9 in. 

 2.06 kip/in.

 2.04 kip/in.

Because there are two beams to be braced, the total required point brace stiffness is: br  2  2.04 kip/in.

LRFD br

 4.08 kip/in.

ASD  2  2.06 kip/in.  4.12 kip/in.

The beams require bracing in order to have sufficient strength to carry the given load. However, locating that brace at the midspan provides flexural strength greater than the required strength. The maximum unbraced length permitted for the required flexural strength is Lb = 18.2 ft from AISC Manual Table 6-2. Thus, according to AISC Specification Appendix 6, Section 6.3.1b, this length could be used in place of 14 ft to determine the required stiffness. However, because the required stiffness is so small, the 14 ft length will be used here. For a single brace, the stiffness is: 

AE cos 2  L

 3.07 in.   29, 000 ksi  5 ft 14.9 ft   2

2

14.9 ft 12 in./ft 

 56.1 kip/in.

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Because there are two braces, the system stiffness is twice this. Thus,   2  56.1 kip/in.  112 kip/in.

LRFD   112 kip/in.  4.08 kip/in. o.k.

ASD   112 kip/in.  4.12 kip/in.

o.k.

Available Strength of Braces The braces may be called upon to act in either tension or compression, depending on which transverse direction the system tries to buckle. Brace compression buckling will control over tension yielding. Therefore, determine the compressive strength of the braces assuming they are eccentrically loaded using AISC Manual Table 4-12. LRFD Interpolating for Lc = 14.9 ft: c Pn  17.2 kips  1.53 kips

ASD Interpolating for Lc = 14.9 ft: o.k.

Pn  11.2 kips  1.03 kips c

o.k.

The L55c braces have sufficient strength and stiffness to act as the point braces for this system.

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EXAMPLE A-6.4 POINT STABILITY BRACING OF A BEAMCASE II Given:

A walkway in an industrial facility has a span of 28 ft as shown in Figure A-6.4-1. The walkway has a deck of grating which is not sufficient to brace the beams. The ASTM A992 W1222 beams are braced against twist at the ends, and they are connected by a strut connected at midspan. At that same point they are braced to an adjacent ASTM A500 Grade C HSS884 column by the attachment of a 5-ft-long ASTM A36 2L334. The brace connections are all welded; therefore, bolt slippage does not need to be accounted for in the stiffness calculation. The adjacent column is not braced at the walkway level, but is adequately braced 12 ft below and 12 ft above the walkway level. The dead load on each beam is 0.05 kip/ft and the live load is 0.125 kip/ft. Determine if the bracing system has adequate strength and stiffness to brace this walkway.

Fig. A-6.4-1. Plan view for Example A-6.4. Solution:

Because the bracing system does not interact directly with any other braced point on the beam, the double angle and column constitute a point brace system. The strut will be assumed to be sufficiently strong and stiff to force the two beams to buckle together. From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi HSS column ASTM A500 Grade C Fy = 50 ksi Fu = 62 ksi Double-angle brace ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-1, 1-12 and 1-15, the geometric properties are as follows:

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Beam W1222

ho = 11.9 in. HSS column HSS884 I = 70.7 in.4 Double-angle brace 2L334 A = 2.88 in.2 Required Flexural Strength of Beam From ASCE/SEI 7, Chapter 2, the required strength is:

LRFD wu  1.2  0.05 kip/ft   1.6  0.125 kip/ft 

ASD wa  0.05 kip/ft  0.125 kip/ft  0.175 kip/ft

 0.260 kip/ft

Determine the required flexural strength for a uniformly distributed load on the simply supported beam using AISC Manual Table 3-23, Case 1, as follows: LRFD Mu  

ASD

2

wu L 8

Ma 

 0.260 kip/ft  28 ft 2



8  25.5 kip-ft

2

wa L 8

 0.175 kip/ft  28 ft 2 8

 17.2 kip-ft

It can be shown that the W1222 beams are adequate with this unbraced length of 14 ft. Both beams need bracing in the same direction simultaneously. Required Brace Strength and Stiffness From AISC Specification Appendix 6, Section 6.3.1b, the required brace force for each beam, with Cd = 1.0 for bending in single curvature, is determined as follows: LRFD

ASD

Mr  Mu

Mr  Ma

 25.5 kip-ft M C  Pbr  0.02  r d   ho 

 17.2 kip-ft (Spec. Eq. A-6-7)

  25.5 kip-ft 12 in. / ft 1.0    0.02   11.9 in.    0.514 kip

M C  Pbr  0.02  r d  (Spec. Eq. A-6-7)  ho   17.2 kip-ft 12 in. / ft 1.0    0.02   11.9 in.    0.347 kip

Because there are two beams, the total required brace force is:

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Pbr  2  0.514 kip 

LRFD

Pbr  2  0.347 kip 

 1.03 kips

ASD

 0.694 kip

By inspection, the 2L334 can carry the required bracing force. The HSS column can also carry the bracing force through bending on a 24-ft-long span. It will be shown that the change in length of the 2L334 is negligible, so the available brace stiffness will come from the flexural stiffness of the column only. From AISC Specification Appendix 6, Section 6.3.1b, with Cd = 1.0 for bending in single curvature, the required brace stiffness is: LRFD

  0.75

  2.00

Mr  Ma

Mr  Mu

 17.2 kip-ft

 25.5 kip-ft br 



ASD

1  10 M r Cd      Lbr ho 

(Spec. Eq. A-6-8a)

1 10  25.5 kip-ft 12 in./ft 1.0     0.75  14 ft 12 in./ft 11.9 in. 

 2.04 kip/in.

 10 M r Cd  br      Lbr ho 

(Spec. Eq. A-6-8b)

10 17.2 kip-ft 12 in./ft 1.0    2.00    14 ft 12 in./ft 11.9 in.   2.06 kip/in.

The beams require one brace in order to have sufficient strength to carry the given load. However, locating that brace at midspan provides flexural strength greater than the required strength. The maximum unbraced length permitted for the required flexural strength is Lb = 18.2 ft from AISC Manual Table 6-2. Thus, according to AISC Specification Appendix 6, Section 6.3.1b, this length could be used in place of 14 ft to determine the required stiffness. Available Stiffness of Brace Because the brace stiffness comes from the combination of the axial stiffness of the double-angle member and the flexural stiffness of the column loaded at its midheight, the individual element stiffness will be determined and then combined. The axial stiffness of the double angle is: 

AE L

 2.88 in.   29, 000 ksi   2

 5 ft 12 in./ft 

 1,390 kip/in.

The available flexural stiffness of the HSS column with a point load at midspan using AISC Manual Table 3-23, Case 7, is:

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 

48 EI L3



48  29, 000 ksi  70.7 in.4

 24.0 ft  12 in./ft  3



3

 4.12 kip/in.

The combined stiffness is: 1 1 1    angles column 1 1  1,390 kip/in. 4.12 kip/in.  0.243 in./kip 

Thus, the system stiffness is:   4.12 kip/in.

The stiffness of the double-angle member could have reasonably been ignored. Because the double-angle brace is ultimately bracing two beams, the required stiffness is multiplied by 2: LRFD 4.12 kip/in.  2  2.04 kip/in.

ASD 4.12 kip/in.  2  2.06 kip/in.

4.12 kip/in.  4.08 kip/in.

4.12 kip/in.  4.12 kip/in.

o.k.

o.k.

The HSS884 column is an adequate brace for the beams. However, if the column also carries an axial force, it must be checked for combined forces.

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EXAMPLE A-6.5 POINT STABILITY BRACING OF A BEAM WITH REVERSE CURVATURE BENDING Given:

A roof system is composed of 26K8 steel joists spaced at 5-ft intervals and supported on ASTM A992 W2150 girders as shown in Figure A-6.5-1(a). The roof dead load is 33 psf and the roof live load is 25 psf. Determine the required strength and stiffness of the braces needed to brace the girder at the support and near the inflection point. Bracing for the beam is shown in Figure A-6.5-1(b). Moment diagrams for the beam are shown in Figures A-6.51(c) and A-6.5-1(d). Determine the size of single-angle kickers connected to the bottom flange of the girder and the top chord of the joist, as shown in Figure A-6.5-1(e), where the brace force will be taken by a connected rigid diaphragm.

(a) Plan

(b) Section B-B: Beam with bracing at top flanges by the steel joists and at the bottom flanges by the single-angle kickers

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(c) Moment diagram of beam

(d) Moment diagram between points B and C

(e) Bracing configuration Fig. A-6.5-1. Example A-6.5 configuration. Solution:

Since the braces will transfer their force to a rigid roof diaphragm, they will be treated as point braces. From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi

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Single-angle brace ASTM A36 Fy = 36 ksi Fu = 58 ksi From the Steel Joist Institute: Joist K-Series Fy = 50 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W2150 ho = 20.3 in. Required Flexural Strength of Beam From ASCE/SEI 7, Chapter 2, the required strength is:

LRFD wu  1.2  33 psf   1.6  25 psf 

 58.0 psf

 79.6 psf wu 

ASD wa  33 psf  25 psf

 79.6 psf  40 ft 

wa 

 58.0 psf  40 ft 

1, 000 lb/kip  2.32 kip/ft

1, 000 lb/kip  3.18 kip/ft

From Figure A-6.5-1(d):

From Figure A-6.5-1(d):

M uB  88.7  3.18 kip/ft 

M aB  88.7  2.32 kip/ft   206 kip-ft

 282 kip-ft Required Brace Strength and Stiffness

Determine the required force to brace the bottom flange of the girder with a point brace. The braces at points B and C will be determined based on the moment at B. However, because the brace at C is the closest to the inflection point, its strength and stiffness requirements are greater since they are influenced by the variable Cd which will be equal to 2.0. From AISC Specification Appendix 6, Section 6.3.1b, the required brace force is determined as follows: LRFD M r  M uB  282 kip-ft

ASD M r  M aB  206 kip-ft

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LRFD M C  Pbr  0.02  r d   ho 

(Spec. Eq. A-6-7)

  282 kip-ft 12 in./ft  2.0    0.02   20.3 in.    6.67 kips

ASD M C  Pbr  0.02  r d   ho 

(Spec. Eq. A-6-7)

  206 kip-ft 12 in./ft  2.0    0.02   20.3 in.    4.87 kips

Determine the required stiffness of the point brace at point C. The required brace stiffness is a function of the unbraced length. It is permitted to use the maximum unbraced length permitted for the beam based upon the required flexural strength. Thus, determine the maximum unbraced length permitted. Based on AISC Specification Section F1 and the moment diagram shown in Figure A-6.5-1(d), for the beam between points B and C, the lateral-torsional buckling modification factor, Cb, is:

Cb  

2.5M max

12.5M max  3M A  4 M B  3M C

(Spec. Eq. F1-1)

12.5  88.7 w 

2.5  88.7 w   3  41.8w   4  1.2w   3  32.2w 

 2.47 The maximum unbraced length for the required flexural strength can be determined by setting the available flexural strength based on AISC Specification Equation F2-3 (lateral-torsional buckling) equal to the required strength and solving for Lb (this is assuming that Lb > Lr). LRFD For a required flexural strength, Mu = 282 kip-ft, with Cb = 2.47, the unbraced length may be taken as:

ASD For a required flexural strength, Ma = 206 kip-ft, with Cb = 2.47, the unbraced length may be taken as:

Lb = 22.0 ft

Lb = 20.6 ft

From AISC Specification Appendix 6, Section 6.3.1b, the required brace stiffness is:   0.75

LRFD

ASD  = 2.00 M r  M aB

M r  M uB  282 kip-ft br 



1  10 M r Cd      Lbr ho 

 206 kip-ft

(Spec. Eq. A-6-8a)

1 10  282 kip-ft 12 in./ft  2.0     0.75   22.0 ft 12 in./ft  20.3 in. 

 16.8 kip/in.

 10 M r Cd  br      Lbr ho 

(Spec. Eq. A-6-8b)

10  206 kip-ft 12 in./ft  2.0    2.00     20.6 ft 12 in./ft  20.3 in.   19.7 kip/in.

Because no deformation will be considered in the connections, only the brace itself will be used to provide the required stiffness. The brace is oriented with the geometry as shown in Figure A-6.5-1(e). Thus, the force in the brace is Fbr = Pbr/(cosθ) and the stiffness of the brace is AE(cos2θ)/L. There are two braces at each brace point. One would be in tension and one in compression, depending on the direction that the girder attempts to buckle. For

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simplicity in design, a single brace will be selected that will be assumed to be in tension. Only the limit state of yielding will be considered. Select a single angle to meet the requirements of strength and stiffness, with a length of:

 48 in.2   20 in.2

L

 52.0 in.

Required Brace Force

LRFD

ASD

Pbr cos  6.67 kips   48.0 in. 52.0 in.

Pbr cos  4.87 kips   48.0 in. 52.0 in.

Fbr 

Fbr 

 7.23 kips

 5.28 kips

From AISC Specification Section D2(a), the required area based on available tensile strength is determined as follows:

Ag  

Fbr Fy

(modified Spec. Eq. D2-1)

7.23 kips 0.90  36 kips 



2

Fbr Fy

Ag 

(modified Spec. Eq. D2-1)

1.67  5.28 kips  36 kips

 0.245 in.2

 0.223 in.

The required area based on stiffness is: LRFD Ag  

br L E cos 2  16.8 kip/in. 52.0 in.

 29,000 ksi  48.0 in. 52.0 in.2

 0.0354 in.2

ASD Ag  

br L E cos 2  19.7 kip/in. 52.0 in.

 29,000 ksi  48.0 in. 52.0 in.2

 0.0415 in.2

The strength requirement controls, therefore select L228 with A = 0.491 in.2 At the column at point B, the required strength would be one-half of that at point C, because Cd = 1.0 at point B instead of 2.0. However, since the smallest angle available has been selected for the brace, there is no reason to check further at the column and the same angle will be used there.

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EXAMPLE A-6.6 POINT TORSIONAL STABILITY BRACING OF A BEAM Given:

A roof system is composed of ASTM A992 W1240 intermediate beams spaced 5 ft on center supporting a connected panel roof system that cannot be used as a diaphragm. As shown in Figure A-6.6-1, the beams span 30 ft and are supported on W3090 girders spanning 60 ft. This is an isolated roof structure with no connections to other structures that could provide lateral support to the girder compression flanges. Thus, the flexural resistance of the attached beams must be used to provide torsional stability bracing of the girders. The roof dead load is 40 psf and the roof live load is 24 psf. Determine if the beams are sufficient to provide point torsional stability bracing.

(a) Plan

(b) Point torsional brace connection Fig. A-6.6-1. Roof system configuration

Solution:

Because the bracing beams are not connected in a way that would permit them to transfer an axial bracing force, they must behave as point torsional braces if they are to effectively brace the girders. From AISC Manual Table 2-4, the material properties are as follows: Beam and girder ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1240

tw = 0.295 in. Ix = 307 in.4 Girder W3090

tw = 0.470 in. ho = 28.9 in. Iy = 115 in.4

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Required Flexural Strength of Girder From ASCE/SEI 7, Chapter 2, and using AISC Manual Table 3-23, Case 1, the required strength of the girder is: LRFD wu  1.2  40 psf   1.6  24 psf 

ASD wa  40 psf  24 psf

 64.0 psf

 86.4 psf wu 

 86.4 psf 15 ft 

wa 

1, 000 lb/kip  0.960 kip/ft

1, 000 lb/kip  1.30 kip/ft

Mu  

 64.0 psf 15 ft 

wu L2 8

Ma 

1.30 kip/ft  60 ft 2



8  585 kip-ft

wa L2 8

 0.960 kip/ft  60 ft 2 8

 432 kip-ft

With Cb = 1.0, from AISC Manual Table 3-10, the maximum unbraced length permitted for the W3090 based upon required flexural strength is: LRFD For MuB = 585 kip-ft, Lb = 22.0 ft

ASD For MaB = 432 kip-ft, Lb = 20.7 ft

Point Torsional Brace Design The required flexural strength for a point torsional brace for the girder is determined from AISC Specification Appendix 6, Section 6.3.2a. LRFD

ASD

M r  M uB

M r  M aB

 585 kip-ft M br  0.02 M r  0.02  585 kip-ft 

 11.7 kip-ft

 432 kip-ft (Spec. Eq. A-6-9)

M br  0.02 M r  0.02  432 kip-ft 

(Spec. Eq. A-6-9)

 8.64 kip-ft

The required overall point torsional brace stiffness with braces every 5 ft, n = 11, and assuming Cb = 1.0, is determined in the following. Based on the User Note in Specification Section 6.3.2a:

I yeff  I y  115 in.4

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LRFD

  0.75

ASD

  3.00

2

1 2.4 L  M r  T  (Spec. Eq. A-6-11a)  nEI yeff  Cb    1  2.4  60 ft 12 in./ft    0.75 11 29, 000 ksi  115 in.4   



2



  585 kip-ft 12 in./ft     1.0    3,100 kip-in./rad

2.4 L  M r  (Spec. Eq. A-6-11b) nEI yeff  Cb   2.4 60 ft 12 in./ft      3.00  11 29, 000 ksi  115 in.4   

T  



  432 kip-ft 12 in./ft     1.0    3,800 kip-in./rad

2



2

The distortional buckling stiffness of the girder web is a function of the web slenderness and the presence of any stiffeners. The web distortional stiffness is:

sec 

3.3E  1.5ho tw3 tst bs3     ho  12 12 

(Spec. Eq. A-6-12)

Therefore the distortional stiffness of the girder web alone is: sec 

3.3E  1.5ho tw3  ho  12

  

3.3  29, 000 ksi  1.5  28.9 in. 0.470 in.  28.9 in. 12   1, 240 kip-in./rad

3



  

For AISC Specification Equation A-6-10 to give a nonnegative result, the web distortional stiffness given by Equation A-6-12 must be greater than the required point torsional stiffness given by Equation A-6-11. Because the web distortional stiffness of the girder is less than the required point torsional stiffness for both LRFD and ASD, web stiffeners will be required. Determine the torsional stiffness contributed by the beams. Both girders will buckle in the same direction forcing the beams to bend in reverse curvature. Thus, the flexural stiffness of the beam using AISC Manual Table 3-23, Case 9, is: Tb  

6 EI L



6  29, 000 ksi  307 in.4

 30 ft 12 in./ft 



 148, 000 kip-in./rad

Determining the required distortional stiffness of the girder will permit determination of the required stiffener size. The total stiffness is determined by summing the inverse of the distortional and flexural stiffnesses. Thus: 1 1 1   T Tb  sec

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Determine the minimum web distortional stiffness required to provide bracing for the girder. LRFD

ASD

1 1 1   T Tb sec 1 1 1   3,100 148, 000 sec

1 1 1   T Tb sec 1 1 1   3,800 148, 000 sec

 sec  3,170 kip-in./rad

 sec  3, 900 kip-in./rad

Determine the required width, bs, of a-in.-thick stiffeners.

 sec 

 1.5ho tw3

3.3E  ho 

12

LRFD t b3   st s  12 

(Spec. Eq. A-6-12)

 sec 

 1.5ho tw3

3.3E  ho 

12

ASD t b3   st s  12 

(Spec. Eq. A-6-12)

Using the total required girder web distortional stiffness and the contribution of the girder web distortional stiffness calculated previously, solve for the required width for a-in.-thick stiffeners:

Using the total required girder web distortional stiffness and the contribution of the girder web distortional stiffness calculated previously, solve for the required width for a-in.-thick stiffeners:

3,170 kip-in./rad  1, 240 kip-in./rad

3,900 kip-in./rad  1, 240 kip-in./rad



3.3(29, 000 ksi)   a in.  28.9 in. 12 

and bs = 2.65 in.

bs3

  



3 3.3(29, 000 ksi)   a in. bs    28.9 in. 12  

and bs = 2.95 in.

Therefore, use a 4 in. x a in. full depth one-sided stiffener at the connection of each beam. Available Flexural Strength of Beam Each beam is connected to a girder web stiffener. Thus, each beam will be coped at the top and bottom as shown in Figure A-6.6-1(b) with a depth at the coped section of 9 in. The available flexural strength of the coped beam is determined using the provisions of AISC Specification Sections J4.5 and F11. M n  M p  Fy Z  1.6 Fy S x

(Spec. Eq. F11-1)

For a rectangle, Z < 1.6S. Therefore, strength will be controlled by FyZ and Z

 0.295 in. 9.00 in.2 4 3

 5.97 in.

The nominal flexural strength of the beam is: M n  Fy Z x

 50 ksi   5.97 in.3   12 in./ft   24.9 kip-ft

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LRFD

ASD

 = 0.90

Ω = 1.67

M n  0.90  24.9 kip-ft 

M n 24.9 kip-ft   1.67  14.9 kip-ft  8.64 kip-ft o.k.

 22.4 kip-ft  11.7 kip-ft

o.k.

Neglecting any rotation due to the bolts moving in the holes or any influence of the end moments on the strength of the beams, this system has sufficient strength and stiffness to provide point torsional bracing to the girders. Additional connection design limit states may also need to be checked.

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APPENDIX 6 REFERENCES

Helwig, Todd A. and Yura, J.A. (1999), “Torsional Bracing of Columns,” Journal of Structural Engineering, ASCE, Vol. 125, No. 5, pp. 547555. Yura, J.A. (2001), “Fundamentals of Beam Bracing,” Engineering Journal, AISC, Vol. 38, No. 1, pp. 1126. Ziemian, R.D. (ed.) (2010), Guide to Stability Design Criteria for Metal Structures, 6th Ed., John Wiley & Sons, Inc., Hoboken, NJ.

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Chapter IIA Simple Shear Connections The design of connecting elements are covered in Part 9 of the AISC Manual. The design of simple shear connections is covered in Part 10 of the AISC Manual.

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EXAMPLE II.A-1A ALL-BOLTED DOUBLE-ANGLE CONNECTION Given: Using the tables in AISC Manual Part 10, verify the available strength of an all-bolted double-angle shear connection between an ASTM A992 W36231 beam and an ASTM A992 W1490 column flange, as shown in Figure IIA-1A-1, supporting the following beam end reactions: RD = 37.5 kips RL = 113 kips Use ASTM A36 angles.

Fig. IIA-1A-1. Connection geometry for Example II.A-1A. Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi

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From AISC Manual Table 1-1, the geometric properties are as follows: Beam W36231

tw = 0.760 in. Column W1490

tf = 0.710 in. From AISC Specification Table J3.3, the hole diameter for a w-in.-diameter bolt with standard holes is: d h  m in.

From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  37.5 kips   1.6 113 kips 

ASD Ra  37.5 kips  113 kips

 151 kips

 226 kips Connection Selection

AISC Manual Table 10-1 includes checks for the limit states of bolt shear, bolt bearing and tearout on the angles, shear yielding of the angles, shear rupture of the angles, and block shear rupture of the angles. Try 8 rows of bolts and 2L532c (SLBB). From AISC Manual Table 10-1: LRFD Rn  248 kips  226 kips

o.k. 

ASD Rn  165 kips  151 kips o.k. 

Available Beam Web Strength The available beam web strength is the lesser of the limit states of block shear rupture, shear yielding, shear rupture, and the sum of the effective strengths of the individual fasteners. Because the beam is not coped, the only applicable limit state is the effective strength of the individual fasteners, which is the lesser of the bolt shear strength per AISC Specification Section J3.6, and the bolt bearing and tearout strength per AISC Specification Section J3.10. Bolt Shear From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear is: LRFD

Rn  35.8 kips/bolt

ASD Rn  23.9 kips/bolt 

Bolt Bearing on Beam Web The nominal bearing strength of the beam web per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration:

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rn  2.4dtFu

(Spec. Eq. J3-6a)

 2.4  w in. 0.760 in. 65 ksi   88.9 kips/bolt

From AISC Specification Section J3.10, the available bearing strength of the beam web per bolt is:   0.75

LRFD

  2.00

rn  0.75  88.9 kips/bolt 

ASD

rn 88.9 kips/bolt   2.00  44.5 kips/bolt

 66.7 kips/bolt Bolt Tearout on Beam Web

The available tearout strength of the beam web per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration: lc  3.00 in.  m in.  2.19 in.

The available tearout strength is:

rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2  2.19 in. 0.760 in. 65 ksi   130 kips/bolt From AISC Specification Section J3.10, the available tearout strength of the beam web per bolt is:   0.75

LRFD

  2.00

rn  0 130 kips/bolt 

ASD

rn 130 kips/bolt     65.0 kips/bolt

 97.5 kips/bolt

Bolt shear strength is the governing limit state for all bolts at the beam web. Bolt shear strength is one of the limit states included in the capacities shown in Table 10-1 as used above; thus, the effective strength of the fasteners is adequate. Available Strength at the Column Flange Since the thickness of the column flange, tf = 0.710 in., is greater than the thickness of the angles, t = c in., bolt bearing will control for the angles, which was previously checked. The column flange is adequate for the required loading. Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-1B ALL-BOLTED DOUBLE-ANGLE CONNECTION SUBJECT TO AXIAL AND SHEAR LOADING Given:

Verify the available strength of an all-bolted double-angle connection for an ASTM A992 W1850 beam, as shown in Figure II.A-1B-1, to support the following beam end reactions: LRFD Shear, Vu = 75 kips Axial tension, Nu = 60 kips

ASD Shear, Va = 50 kips Axial tension, Na = 40 kips

Use ASTM A36 double angles that will be shop-bolted to the beam.

Fig. II.A-1B-1. Connection geometry for Example II.A-1B. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi

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From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1850 Ag = 14.7 in.2 d = 18.0 in. tw = 0.355 in. tf = 0.570 in. From AISC Specification Table J3.3, the hole diameter for d-in.-diameter bolts with standard holes is: dh = , in. The resultant load is: LRFD 2

Ru  Vu  N u 

ASD

2

 75 kips 

2

2

Ra  Va  N a   60 kips 

2



 96.0 kips

2

 50 kips 2   40 kips 2

 64.0 kips

Try 5 rows of bolts and 2L532s (SLBB). Strength of the Bolted Connection—Angles From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the invividual strengths of the individual fasteners, which may be taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. Bolt shear From AISC Manual Table 7-1, the available shear strength for d-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear (or pair of bolts) is: LRFD

rn  48.7 kips/bolt (or per pair of bolts)

ASD rn  32.5 kips/bolt (or per pair of bolts) 

Bolt bearing on angles The available bearing strength of the angles per bolt in double shear is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration:

rn   2 angles  2.4dtFu

(from Spec. Eq. J3-6a)

  2 angles  2.4  d in. s in. 58 ksi   152 kips/bolt

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LRFD

  0.75

ASD

  2.00

rn 152 kips/bolt     76.0 kips/bolt

rn  0.75 152 kips/bolt   114 kips/bolt Bolt tearout on angles

From AISC Specification Section J3.10, the available tearout strength of the angles per bolt in double shear is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration. As shown in Figures II.A-1B-2(a) and II.A-1B-2(b), the tearout dimensions on the angle differ between the edge bolt and the other bolts. The angle , as shown in Figure II.A-1B-2(a), of the resultant force on the edge bolt is: LRFD

ASD

N    tan 1  u   Vu 

N    tan 1  a   Va 

 60 kips   tan 1    75 kips   38.7

 40 kips   tan 1    50 kips   38.7

         

 

          (a) Edge bolt

(b) Other bolts

Fig. II.A-1B-2. Bolt tearout on angles.

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The length from the center of the bolt hole to the edge of the angle along the line of action of the force is:

14 in. cos 38.7  1.60 in.

le 

The clear distance, along the line of action of the force, between the edge of the hole and the edge of the angle is:

lc  le  0.5d h  1.60 in.  0.5 , in.  1.13 in. The available tearout strength of the pair of angles at the edge bolt is: rn   2 angles 1.2lc tFu

(from Spec. Eq. J3-6c)

  2 angles 1.2 1.13 in. s in. 58 ksi   98.3 kips/bolt

  0.75

LRFD

rn  0  98.3 kips/bolt   73.7 kips/bolt

  2.00

ASD

rn 98.3 kips/bolt     49.2 kips/bolt

Therefore, bolt shear controls over bearing or tearout of the angles at the edge bolt. The angle as shown in Figure II.A-1B-2(b), of the resultant force on the other bolts is: LRFD V    tan 1  u   Nu   75 kips   tan 1    60 kips   51.3

ASD V    tan 1  a   Na   50 kips   tan 1    40 kips   51.3

The length from the center of the bolt hole to the edge of the angle along the line of action of the force is:

14 in. cos 51.3  2.00 in.

le 

The clear distance, along the line of action of the force, between the edge of the hole and the edge of the angle is:

lc  le  0.5d h  2.00 in.  0.5 , in.  1.53 in. The available tearout strength of the pair of angles at the other bolts is:

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rn   2 angles 1.2lc tFu

(from Spec. Eq. J3-6c)

  2 angles 1.2 1.53 in. s in. 58 ksi   133 kips/bolt

  0.75

LRFD

  2.00

rn  0 133 kips/bolt 

ASD

rn 133 kips/bolt     66.5 kips/bolt

 99.8 kips/bolt

Therefore, bolt shear controls over bearing or tearout of the angles at the other bolt. The effective strength for the bolted connection at the angles is determined by summing the effective strength for each bolt using the minimum available strength calculated for bolt shear, bearing on the angles, and tearout on the angles. LRFD

ASD

  5 bolts  48.7 kips/bolt 

Rn r n n     5 bolts  32.5 kips/bolt 

Rn  nrn  244 kips  96.0 kips o.k.

 163 kips  64.0 kips o.k.

Strength of the Bolted Connection—Beam Web Bolt bearing on beam web The available bearing strength of the beam web per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration: rn  2.4dtFu

(Spec. Eq. J3-6a)

 2.4  d in. 0.355 in. 65 ksi   48.5 kips/bolt   0.75

LRFD

rn  0.75  48.5 kips/bolt   36.4 kips/bolt

  2.00

ASD

rn 48.5 kips/bolt  2.00   24.3 kips/bolt

Bolt tearout on beam web From AISC Specification Section J3.10, the available tearout strength of the beam web is determined from AISC Specification Equation J3-6a, assuming deformation at the bolt hole is a design consideration, where the edge distance, lc, is based on the angle of the resultant load. As shown in Figure II.A-1B-3, a horizontal edge distance of 12 in. is used which includes a 4 in. tolerance to account for possible mill underrun. The angle, , of the resultant force is:

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LRFD

ASD

V    tan 1  u   Nu 

V    tan 1  a   Na   50 kips   tan 1    40 kips 

 75 kips   tan 1    60 kips   51.3

 51.3

The length from the center of the bolt hole to the edge of the web along the line of action of the force is: 12 in. cos 51.3  2.40 in.

le 

The clear distance, along the line of action of the force, between the edge of the hole and the edge of the web is:

lc  le  0.5d h  2.40 in.  0.5 , in.  1.93 in. The available tearout strength of the beam web is determined as follows:

rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2 1.93 in. 0.355 in. 65 ksi   53.4 kips/bolt

  0.75

LRFD

rn  0  53.4 kips/bolt   40.1 kips/bolt

  2.00

rn 53.4 kips/bolt     26.7 kips/bolt

Fig. II.A-1B-3. Bolt tearout on beam web.

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Therefore, bolt bearing on the beam web is the controlling limit state for all bolts. The effective strength for the bolted connection at the beam web is determined by summing the effective strength for each bolt using the minimum available strength calculated for bolt shear, bearing on the beam web, and tearout on the beam web. LRFD

ASD Rn rn n     5 bolts  24.3 kips/bolt 

Rn  nrn   5 bolts  36.4 kips/bolt   182 kips  96.0 kips o.k.

 122 kips  64.0 kips o.k.

Bolt Shear and Tension Interaction—Outstanding Angle Legs The available tensile strength of the bolts due to the effect of combined tension and shear is determined from AISC Specification Section J3.7. The required shear stress is:

f rv 

Vr nAb

where Ab  0.601 in.2 (from AISC Manual Table 7-1)

n  10 LRFD f rv

ASD

V  u nAb 



f rv 75 kips 2

10 0.601 in.

V  a nAb 



 12.5 ksi



50 kips

10 0.601 in.2



 8.32 ksi

The nominal tensile strength modified to include the effects of shear stress is determined from AISC Specification Section J3.7 as follows. From AISC Specification Table J3.2:

Fnt  90 ksi Fnv  54 ksi   0.75

LRFD

  2.00

Fnt f rv  Fnt (Spec. Eq. J3-3a) Fnv 90 ksi  1.3  90 ksi   12.5 ksi   90 ksi 0.75  54 ksi 

Fnt  1.3Fnt 

 89.2 ksi  90 ksi

ASD

Fnt f rv  Fnt (Spec. Eq. J3-3b) Fnv 2.00  90 ksi   1.3  90 ksi   8.32 ksi   90 ksi 54 ksi  89.3 ksi  90 ksi

Fnt  1.3Fnt 

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LRFD

ASD

Therefore:

Therefore:

Fnt  89.2 ksi

Fnt  89.3 ksi

Using the value of Fnt determined for LRFD, the nominal tensile strength of one bolt is:

rn  Fnt Ab



  89.2 ksi  0.601 in.2

(from Spec. Eq. J3-2)



 53.6 kips The available tensile strength of the bolts due to combined tension and shear is: LRFD

  0.75

  2.00

rn  0.75  53.6 kips/bolt 

rn 53.6 kips/bolt   2.00  26.8 kips

 40.2 kips

Rn r n n    10 bolts  26.8 kips/bolt 

Rn  nrn  10 bolts  40.2 kips/bolt   402 kips  60 kips

ASD

o.k.

 268 kips  40 kips o.k.

Prying Action From AISC Manual Part 9, the available tensile strength of the bolts in the outstanding angle legs taking prying action into account is determined as follows: a 

2( angle leg )  t w  gage 2 2  5 in.  0.355 in.  72 in. 2

 1.43 in.

gage  tw  t 2 72 in.  0.355 in.  s in.  2  3.26 in.

b

d   d   a    a  b   1.25b  b  2   2   d in. d in.  1.43 in.   1.25  3.26 in.  2 2  1.87 in.  4.51 in. o.k.

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d   b   b  b  2    3.26 in. 

(Manual Eq. 9-18) d in. 2

 2.82 in. b a 2.82 in.  1.87 in.  1.51



(Manual Eq. 9-22)

Note that end distances of 14 in. are used on the angles, so p is the average pitch of the bolts: l n 142 in.  5 rows  2.90 in.

p

Check: p  s  3.00 in.

o.k.

d p , in.  1 2.90 in.  0.677

  1

(Manual Eq. 9-20)

The angle thickness required to develop the available strength of the bolt with no prying action is determined as follows:   0.90

LRFD

Bc  40.2 kips/bolt (calculated previously)

tc  

4 Bc b pFu 4  40.2 kips/bolt  2.82 in. 0.90  2.90 in. 58 ksi 

 1.73 in.

ASD

  1.67

(Manual Eq. 9-26a)

Bc  26.8 kips/bolt (calculated previously)

tc  

4 Bc b pFu

(Manual Eq. 9-26b)

1.67  4  26.8 kips/bolt  2.82 in.

 2.90 in. 58 ksi 

 1.73 in.

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2  1   tc     1 (1  )  t    1.73 in.  2  1     1 0.677 1   s in.    3.92

 

(Manual Eq. 9-28)

Because    1, the angles have insufficient strength to develop the bolt strength, therefore: 2

t  Q    1     tc  2

 s in.    1     1.73 in.   0.219

The available tensile strength of the bolts, taking prying action into account, is determined using AISC Manual Equation 9-27, as follows: LRFD rn  Bc Q   40.2 kips/bolt  0.219   8.80 kips/bolt

ASD rn  Bc Q    26.8 kips/bolt  0.219 

 5.87 kips/bolt Rn  nrn  10 bolts  8.80 kips/bolt   88.0 kips  60 kips

o.k.

Rn r n n    10 bolts  5.87 kips/bolt   58.7 kips  40 kips

o.k.

Shear Strength of Angles From AISC Specification Section J4.2(a), the available shear yielding strength of the angles is determined as follows: Agv   2 angles  lt   2 angles 142 in. s in.  18.1 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  18.1 in.

2



 391 kips

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LRFD

  1.00

  1.50

Rn  1.00  391 kips 

ASD

Rn 391 kips  1.50   261 kips  64.0 kips o.k.

 391 kips  96.0 kips o.k.

From AISC Specification Section J4.2, the available shear rupture strength of the angle is determined using the net area determined in accordance with AISC Specification Section B4.3b. Anv   2 angles  l  n  d h  z in.  t   2 angles  142 in.  5 , in.  z in.   s in.  11.9 in.2

Rn  0.60 Fu Anv



 0.60  58 ksi  11.9 in.

2

(Spec. Eq. J4-4)



 414 kips LRFD

  0.75

Rn  0.75  414 kips   311 kips  96.0 kips o.k.

  2.00

ASD

Rn 414 kips  2.00   207 kips  64.0 kips o.k.

Tensile Strength of Angles From AISC Specification Section J4.1(a), the available tensile yielding strength of the angles is determined as follows: Ag   2 angles  lt   2 angles 142 in. s in.  18.1 in.2 Rn  Fy Ag

(Spec. Eq. J4-1)



  36 ksi  18.1 in.

2



 652 kips

  0.90

LRFD

Rn  0.90  652 kips   587 kips  60 kips

o.k.

  1.67

Rn 652 kips   1.67  390 kips  40 kips

ASD

o.k.

From AISC Specification Sections J4.1, the available tensile rupture strength of the angles is determined from AISC Specification Equation J4-2. Table D3.1, Case 1 applies in this case because the tension load is transmitted directly

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to the cross-sectional element by fasteners; therefore, U = 1.00. With Ant = Anv (calculated previously), the effective net area is:

Ae  AntU



2

 11.9 in.

(Spec. Eq. D3-1)

 1.00

 11.9 in.2 Rn  Fu Ae



  58 ksi  11.9 in.

2

(Spec. Eq. J4-2)



 690 kips   0.75

LRFD

  2.00

Rn  0.75  690 kips   518 kips  60 kips

Rn 690 kips   2.00  345 kips  40 kips

o.k.

ASD

o.k.

Block Shear Rupture of Angles—Beam Web Side The nominal strength for the limit state of block shear rupture of the angles, assuming an L-shaped tearout due the shear load only, is determined as follows. The tearout pattern is shown in Figure II.A-1B-4.

Rbsv  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant where Agv   2 angles  l  lev  t   2 angles 142 in.  14 in. s in.  16.6 in.2

Fig. II.A-1B-4. Block shear rupture of angles for shear load only.

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(Spec. Eq. J4-5)

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Anv  Agv   2 angles  n  0.5  d h  z in. t  16.6 in.2   2 angles  5  0.5 , in.  z in. s in.  11.0 in.2 Ant   2 angles  leh  0.5  d h  z in.  t   2 angles  14 in.  0.5 , in.  z in.   s in.  0.938 in.2 U bs  1.0

and















Rbsv  0.60  58 ksi  11.0 in.2  1.0  58 ksi  0.938 in.2  0.60  36 ksi  16.6 in.2  1.0  58 ksi  0.938 in.2



 437 kips  413 kips

Therefore: Rbsv  413 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture on the angles is: LRFD

  0.75

Rbsv  0.75  413 kips   310 kips  75 kips o.k.

  2.00

ASD

Rbsv 413 kips   2.00  207 kips  50 kips o.k.

The block shear rupture failure path due to axial load only could occur as an L- or U-shape. Assuming an L-shaped tearout relative to the axial load on the angles, the nominal block shear rupture strength in the angles is determined as follows. The tearout pattern is shown in Figure II.A-1B-5.

Fig. II.A-1B-5. Block shear rupture of angles for axial load only—L-shape.

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Rbsn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Agv   2 angles  leh t   2 angles 14 in. s in.  1.56 in.2 Anv  Agv   2 angles  0.5  d h  z in. t  1.56 in.2   2 angles  0.5 , in.  z in. s in.  0.935 in.2 Ant   2 angles   l  lev    n  0.5  d h  z in.  t   2 angles  142 in.  14 in.   5  0.5 , in.  z in.   s in.  10.9 in.2 U bs  1.0

and















Rbsn  0.60  58 ksi  0.935 in.2  1.0  58 ksi  10.9 in.2  0.60  36 ksi  1.56 in.2  1.0  58 ksi  10.9 in.2



 665 kips  666 kips

Therefore: Rbsn  665 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture on the angles is:   0.75

LRFD

  2.00

ASD

Rbsn 665 kips   2.00  333 kips  40 kips o.k.

Rbsn  0.75  665 kips   499 kips  60 kips o.k.

The nominal strength for the limit state of block shear rupture assuming an U-shaped tearout relative to the axial load on the angles is determined as follows. The tearout pattern is shown in Figure II.A-1B-6.

Rbsn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant where Agv   2 angles  2 planes  leh t   2 angles  2 planes 14 in. s in.  3.13 in.2

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(Spec. Eq. J4-5)

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Anv   2 angles  2 planes  leh  0.5  d h  z in.  t   2 angles  2 planes  14 in.  0.5 , in.+z in.   s in.  1.88 in.2 Ant   2 angles  12.0in.   n  1 d h  z in.  t   2 angles  12.0 in.   5  1, in.  z in.   s in.  10.0 in.2

Ubs = 1.0 and















Rbsn  0.60  58 ksi  1.88 in.2  1.0  58 ksi  10.0 in.2  0.60  36 ksi  3.13 in.2  1.0  58 ksi  10.0 in.2



 645 kips  648 kips

Therefore: Rbsn  645 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture on the angles is: LRFD

  0.75

Rbsn  0.75  645 kips   484 kips  60 kips o.k.

  2.00

ASD

Rbsn 645 kips   2.00  323 kips  40 kips o.k.

Fig. II.A-1B-6. Block shear rupture of angles for axial load only—U-shape.

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Considering the interaction of shear and axial loads, apply a formulation that is similar to AISC Manual Equation 10-5: LRFD 2

ASD 2

2

 Vu   Nu      1  Rbsv    Rbsn  

 Vu   Nu      1  Rbsv   Rbsn  2

2

2

 75 kips   60 kips       0.0739  1 o.k.  310 kips   484 kips 

2

2

 50 kips   40 kips       0.0737  1 o.k.  207 kips   323 kips 

Block Shear Rupture of Angles–Outstanding Legs The nominal strength for the limit state of block shear rupture relative to the shear load on the angles is determined as follows. The tearout pattern is shown in Figure II.A-1B-7.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant where Agv   2 angles  l  lev  t   2 angles 142 in.  14 in. s in.  16.6 in.2

Anv  Agv   2 angles  n  0.5  d h  z in. t  16.6 in.2   2 angles  5  0.5 , in.  z in. s in.  11.0 in.2 Ant   2 angles  leh  0.5  d h  z in.  t   2 angles  1v in.  0.5 , in.  z in.   s in.  1.17 in.2

Fig. II.A-1B-7. Block shear rupture of outstanding legs of angles.

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(Spec. Eq. J4-5)

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Back IIA-21

U bs  1.0

and















Rn  0.60  58 ksi  11.0 in.2  1.0  58 ksi  1.17 in.2  0.60  36 ksi  16.6 in.2  1.0  58 ksi  1.17 in.2



 451 kips  426 kips

Therefore: Rn  426 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture on the angles is: LRFD

  0.75

Rn  0.75  426 kips   320 kips  75 kips o.k.

  2.00

ASD

Rn 426 kips   2.00  213 kips  50 kips o.k.

Shear Strength of Beam Web From AISC Specification Section J4.2(a), the available shear yield strength of the beam web is determined as follows: Agv  dtw  18.0 in. 0.355 in.  6.39 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  6.39 in.

2



 192 kips

  1.00

LRFD

Rn  1.00 192 kips   192 kips  75 kips

o.k.

  1.50

Rn 192 kips   1.50  128 kips  50 kips

ASD

o.k.

The limit state of shear rupture of the beam web does not apply in this example because the beam is uncoped. Tensile Strength of Beam From AISC Specification Section J4.1(a), the available tensile yielding strength of the beam is determined as follows:

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Rn  Fy Ag

(Spec. Eq. J4-1)



  50 ksi  14.7 in.2



 735 kips

LRFD

  0.90

  1.67

Rn  0.90  735 kips   662 kips  60 kips

Rn 735 kips   1.67  440 kips  40 kips

o.k.

ASD

o.k.

From AISC Specification Section J4.1(b), determine the available tensile rupture strength of the beam. The effective net area is Ae = AnU. No cases in AISC Specification Table D3.1 apply to this configuration; therefore, U is determined from AISC Specification Section D3. An  Ag  n  d h  z in. tw   14.7 in.2  5 , in.  z in. 0.355 in.  12.9 in.2

As stated in AISC Specification Section D3, the value of U can be determined as the ratio of the gross area of the connected element (beam web) to the member gross area. U

 d  2t f   tw  Ag

18.0 in.  2  0.570 in.   0.355 in.  14.7 in.2  0.407 Ae  AnU



 12.9 in.2

(Spec. Eq. D3-1)

  0.407 

 5.25 in.2

Rn  Fu Ae



  65 ksi  5.25 in.2

(Spec. Eq. J4-2)



 341 kips   0.75

LRFD

  2.00

Rn  0.75  341 kips   256 kips  60 kips

Rn 341 kips   2.00  171 kips  40 kips

o.k.

ASD

o.k.

Block Shear Rupture Strength of Beam Web Block shear rupture is only applicable in the direction of the axial load, because the beam is uncoped and the limit state is not applicable for an uncoped beam subject to vertical shear. Assuming a U-shaped tearout relative to the

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axial load, and assuming a horizontal edge distance of leh = 1w in.  4 in. = 12 in. to account for a possible beam underrun of 4 in., the block shear rupture strength is:

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Agv   2  leh tw   2 12 in. 0.355 in.  1.07 in.2

Anv  Agv   2  0.5  d h  z in. tw  1.07 in.2   2  0.5, in.  z in. 0.355 in.  0.715 in.2

Ant  12.0 in.   n  1 dh  z in.  tw  12.0 in.   5  1, in.  z in.   0.355 in.  2.84 in.2 U bs  1.0

and















Rn  0.60  65 ksi  0.710 in.2  1.0  65 ksi  2.84 in.2  0.60  50 ksi  1.07 in.2  1.0  65 ksi  2.84 in.2



 212 kips  217 kips

Therefore: Rn  212 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture of the beam web is:   0.75

LRFD

Rn  0.75  212 kips   159 kips  60 kips o.k.

  2.00 

ASD

Rn 212 kips   2.00  106 kips  40 kips o.k.

Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-1C ALL-BOLTED DOUBLE-ANGLE CONNECTION—STRUCTURAL INTEGRITY CHECK Given: Verify the all-bolted double-angle connection from Example II.A-1B, as shown in Figure II.A-1C-1, for the structural integrity provisions of AISC Specification Section B3.9. The connection is verified as a beam and girder end connection and as an end connection of a member bracing a column. Note that these checks are necessary when design for structural integrity is required by the applicable building code. The beam is an ASTM A992 W1850 and the angles are ASTM A36 material.

Fig. II.A-1C-1. Connection geometry for Example II.A-1C.

Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Angle ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W18x50

tw = 0.355 in.

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From AISC Specification Table J3.3, the hole diameter for d-in.-diameter bolts with standard holes is: dh = , in. Beam and Girder End Connection From Example II.A-1B, the required shear strength is: LRFD

ASD

Vu  75 kips

Va  50 kips

From AISC Specification Section B3.9(b), the required axial tensile strength is: LRFD 2 Tu  Vu  10 kips 3 2   75 kips   10 kips 3  50 kips  10 kips

ASD Ta  Va  10 kips  50 kips  10 kips

Therefore:

Therefore:

Tu  50 kips

Ta  50 kips

From AISC Specification Section B3.9, these strength requirements are evaluated independently from other strength requirements. Bolt Shear From AISC Specification Section J3.6, the nominal bolt shear strength is: Fnv = 54 ksi, from AISC Specification Table J3.2 Tn  nFnv Ab  2 shear planes 



(from Spec. Eq. J3-1)



  5 bolts  54 ksi  0.601 in.2  2 shear planes   325 kips Bolt Tension From AISC Specification Section J3.6, the nominal bolt tensile strength is: Fnt = 90 ksi, from AISC Specification Table J3.2

Tn  nFnt Ab



 10 bolts  90 ksi  0.601 in.2

(from Spec. Eq. J3-1)



 541 kips Bolt Bearing and Tearout From AISC Specification Section B3.9, for the purpose of satisfying structural integrity requirements, inelastic deformations of the connection are permitted; therefore, AISC Specification Equations J3-6b and J3-6d are used to

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determine the nominal bearing and tearout strength. By inspection the beam web will control. For bolt bearing on the beam web: Tn   5 bolts  3.0dt w Fu

(from Spec. Eq. J3-6b)

  5 bolts  3.0  d in. 0.355 in. 65 ksi   303 kips

For bolt tearout on the beam web (including a 4-in. tolerance to account for possible beam underrun):

lc  leh  0.5d h  1w in.  4 in.  0.5 , in.  1.03 in. Tn   5 bolts 1.5lc tw Fu

(from Spec. Eq. J3-6d)

  5 bolts 1.5 1.03 in. 0.355 in. 65 ksi   178 kips

Angle Bending and Prying Action From AISC Manual Part 9, the nominal strength of the angles accounting for prying action is determined as follows: a 

2( angle leg )  t w  gage 2 2  5 in.  0.355 in.  72 in. 2

 1.43 in.

gage  tw  t 2 72 in.  0.355 in.  s in.  2  3.26 in.

b

db d  1.25b  b 2 2 d in. d in.  1.43 in.   1.25  3.26 in.  2 2  1.87 in.  4.51 in.  1.87 in.

a  a 

(Manual Eq. 9-23)

d   b   b  b  2  

(Manual Eq. 9-18)

 3.26 in. 

d in. 2

 2.82 in.

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b a 2.82 in.  1.87 in.  1.51



(Manual Eq. 9-22)

Note that end distances of 14 in. are used on the angles, so p is the average pitch of the bolts: l n 142 in.  5 bolts  2.90 in.

p

Check: p  s  3.00 in.

o.k.

d   dh  , in.

d p , in.  1 2.90 in.  0.677

  1

Bn  Fnt Ab



  90 ksi  0.601 in.2

(Manual Eq. 9-20)



 54.1 kips/bolt tc  

4 Bn b pFu

(from Manual Eq. 9-26)

4  54.1 kips/bolt  2.82 in.

 2.90 in. 58 ksi 

 1.90 in.   tc  2  1    1  1     t    1.90 in. 2  1     1 0.677 1  1.51  s in.  

 

(Manual Eq. 9-28)

 4.85

Because    1, the angles have insufficient strength to develop the bolt strength, therefore:

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2

t  Q    1     tc  2

 s in.    1  0.677   1.90 in.   0.181

Tn  nBn Q

(from Manual Eq. 9-27)

 10 bolts  54.1 kips/bolt  0.181  97.9 kips

Note: The 97.9 kips includes any prying forces so there is no need to calculate the prying force per bolt, qr. Tensile Yielding of Angles From AISC Specification Section J4.1, the nominal tensile yielding strength of the angles is determined as follows:

Ag   2 angles  lt   2 angles 142 in. s in.  18.1 in.2 Tn  Fy Ag

(from Spec. Eq. J4-1)



  36 ksi  18.1 in.2



 652 kips

Tensile Rupture of Angles From AISC Specification Section J4.1, the nominal tensile rupture strength of the angles is determined as follows: An   2 angles  l  n  d h  z in.  t   2 angles  142 in.  5 , in.  z in.   s in.  11.9 in.2

AISC Specification Table D3.1, Case 1 applies in this case because tension load is transmitted directly to the crosssection element by fasteners; therefore, U = 1.0. Ae  AnU



2

 11.9 in.

(Spec. Eq. D3-1)

 1.0 

 11.9 in.2

Tn  Fu Ae



  58 ksi  11.9 in.

2

(from Spec. Eq. J4-2)



 690 kips

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Block Shear Rupture By inspection, block shear rupture of the beam web will control. From AISC Specification Section J4.3, the available block shear rupture strength of the beam web is determined as follows (account for possible 4-in. beam underrun): Tn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(from Spec. Eq. J4-5)

where Agv  2leh tw  2 1w in.  4 in. 0.355 in.  1.07 in.2 Anv  2 leh  0.5  d h  z in.  tw  2 1w in.  4 in.  0.5 , in.  z in.   0.355 in.  0.710 in.2 Ant  12.0 in.  4  d h  z in.  tw  12.0 in.  4 , in.  z in.   0.355 in.  2.84 in.2 U bs  1.0

and















Tn  0.60  65 ksi  0.710 in.2  1.0  65 ksi  2.84 in.2  0.60  50 ksi  1.07 in.2  1.0  65 ksi  2.84 in.2



 212 kips  217 kips Therefore: Tn  212 kips

Nominal Tensile Strength The controlling nominal tensile strength, Tn, is the least of those previously calculated: Tn  min 325 kips, 541 kips, 97.9 kips, 652 kips, 690 kips, 212 kips  97.9 kips LRFD Tn  97.9 kips  50 kips o.k.

ASD Tn  97.9 kips  50 kips o.k.

Column Bracing From AISC Specification Section B3.9(c), the minimum nominal tensile strength for the connection of a member bracing a column is equal to 1% of two-thirds of the required column axial strength for LRFD and equal to 1% of the required column axial for ASD. These requirements are evaluated independently from other strength requirements. The maximum column axial force this connection is able to brace is determined as follows:

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LRFD

ASD

2 Tn  0.01  Pu 3

Tn  0.01Pa

Solving for the column axial force:

Solving for the column axial force:

3 Pu  100   Tn 2 3  100    97.9 kips  2  14, 700 kips

Pa  100Tn  100  97.9 kips   9, 790 kips

As long as the required column axial strength is less than Pu = 14,700 kips or Pa = 9,790 kips, this connection is an adequate column brace.

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EXAMPLE II.A-2A BOLTED/WELDED DOUBLE-ANGLE CONNECTION Given: Using the tables in AISC Manual Part 10, verify the available strength of a double-angle shear connection with welds in the support legs (welds B) and bolts in the supported-beam-web legs, as shown in Figure II.A-2A-1. The ASTM A992 W36231 beam is attached to an ASTM A992 W1490 column flange supporting the following beam end reactions: RD = 37.5 kips RL = 113 kips Use ASTM A36 angles and 70-ksi weld electrodes.

Fig. II.A-2A-1. Connection geometry for Example II.A-2A. Note: Bottom flange coped for erection.

Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Beam W36231 tw = 0.760 in. Column W1490 tf = 0.710 in. From AISC Specification Table J3.3, the hole diameter for w-in.-diameter bolts with standard holes is: dh = m in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  37.5 kips   1.6 113 kips 

ASD Ra  37.5 kips  113 kips  151 kips

 226 kips Weld Design

Use AISC Manual Table 10-2 (welds B) with n = 8. Try c-in. weld size, l = 232 in. From AISC Manual Table 10-2, the minimum support thickness is: tmin = 0.238 in. < 0.710 in. o.k. LRFD

ASD Rn  186 kips > 151 kips o.k. 

Rn  279 kips > 226 kips o.k.  Angle Thickness

From AISC Specification Section J2.2b, the minimum angle thickness for a c-in. fillet weld is: t  w  z in.  c in.  z in.  a in.

Try 2L432a (SLBB). Angle and Bolt Design AISC Manual Table 10-1 includes checks for bolt shear, bolt bearing and tearout on the angles, shear yielding of the angles, shear rupture of the angles, and block shear rupture of the angles. Check 8 rows of bolts and a-in. angle thickness. LRFD

Rn  284 kips > 226 kips o.k. 

ASD Rn  189 kips > 151 kips o.k. 

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Beam Web Strength The available beam web strength is the lesser of the limit states of block shear rupture, shear yielding, shear rupture, and the sum of the effective strengths of the individual fasteners. In this example, because of the relative size of the cope to the overall beam size, the coped section will not control, therefore, the strength of the bolt group will control (When this cannot be determined by inspection, see AISC Manual Part 9 for the design of the coped section). From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the effective strengths of the individual fasterners. The effective strength of an individual fastener is the lesser of the shear strength, the bearing strength at the bolt holes, and the tearout strength at the bolt holes. Bolt Shear From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear is: LRFD

ASD Rn  23.9 kips/bolt 

Rn  35.8 kips/bolt Bolt Bearing on Beam Web

The nominal bearing strength of the beam web per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration: rn  2.4dtFu

(Spec. Eq. J3-6a)

 2.4  w in. 0.760 in. 65 ksi   88.9 kips/bolt

From AISC Specification Section J3.10, the available bearing strength of the beam web per bolt is:   0.75

LRFD

rn  0.75  88.9 kips/bolt   66.7 kips/bolt

  2.00

ASD

rn 88.9 kips/bolt   2.00  44.5 kips/bolt

Bolt Tearout on Beam Web The available tearout strength of the beam web per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration: lc  3.00 in.  m in.  2.19 in.

rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2  2.19 in. 0.760 in. 65 ksi   130 kips/bolt From AISC Specification Section J3.10, the available tearout strength of the beam web per bolt is:

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  0.75

LRFD

  2.00

rn  0 130 kips/bolt 

ASD

rn 130 kips/bolt     65.0 kips/bolt

 97.5 kips/bolt

Bolt shear strength is the governing limit state for all bolts at the beam web. Bolt shear strength is one of the limit states included in the capacities shown in Table 10-1 as used above; thus, the effective strength of the fasteners is adequate. Available strength at the column flange Since the thickness of the column flange, tf = 0.710 in., is greater than the thickness of the angles, t = a in., shear will control for the angles. The column flange is adequate for the required loading. Summary The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-2B BOLTED/WELDED DOUBLE-ANGLE CONNECTION SUBJECT TO AXIAL AND SHEAR LOADING Given: Verify the available strength of a double-angle connection with welds in the supported-beam-web legs and bolts in the outstanding legs for an ASTM A992 W1850 beam, as showin in Figure II.A-2B-1, to support the following beam end reactions: LRFD Shear, Vu = 75 kips Axial tension, Nu = 60 kips

ASD Shear, Va = 50 kips Axial tension, Na = 40 kips

Use ASTM A36 angles and 70-ksi electrodes.

Fig. II.A-2B-1. Connection geometry for Example II.A-2B.

Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Beam W1850 Ag = 14.7 in.2 d = 18.0 in. tw = 0.355 in. bf = 7.50 in. tf = 0.570 in. From AISC Specification Table J3.3, the hole diameter for d-in.-diameter bolts with standard holes is: dh = , in. The resultant load is: LRFD

ASD

Ru  Vu 2  N u 2 

Ra  Va 2  N a 2

 75 kips 2   60 kips 2



 96.0 kips

 50 kips 2   40 kips 2

 64.0 kips

The following bolt shear, bearing and tearout calculations are for a pair of bolts. Bolt Shear From AISC Manual Table 7-1, the available shear strength for d-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear (or pair of bolts): LRFD

ASD rn  32.5 kips (for pair of bolts) 

rn  48.7 kips (for pair of bolts)

Bolt Bearing on Angles The available bearing strength of the double angle is determined from AISC Specification Section J3.10, assuming deformation at the bolt hole is a design consideration: rn   2 bolts  2.4dtFu

(from Spec. Eq. J3-6a)

  2 bolts  2.4  d in.2 in. 58 ksi   122 kips (for pair of bolts) The available bearing strength for a pair of bolts is:   0.75

LRFD

  2.00

rn  0.75 122 kips 

ASD

rn 122 kips   2.00  61.0 kips (for pair of bolts)

 91.5 kips (for pair of bolts)

The bolt shear strength controls over bearing in the angles.

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Bolt Tearout on Angles The available tearout strength of the angle is determined from AISC Specification Section J3.10, assuming deformation at the bolt hole is a design consideration: For the edge bolt: lc  le  0.5d h  14 in.  0.5 , in.  0.781 in.

rn   2 bolts 1.2lc tFu

(from Spec. Eq. J3-6c)

  2 bolts 1.2  0.781 in.2 in. 58 ksi   54.4 kips (for pair of bolts)

The available tearout strength of the angles for a pair of edge bolts is:   0.75

LRFD

  2.00

rn  0.75  54.4 kips 

ASD

rn 54.4 kips   2.00  27.2 kips

 40.8 kips

The tearout strength controls over bolt shear and bearing for the edge bolts in the angles. For the other bolts:

lc  s  dh  3 in.  , in.  2.06 in. rn   2 bolts 1.2lc tFu

(Spec. Eq. J3-6c)

  2 bolts 1.2  2.06 in.2 in. 58 ksi   143 kips (for pair of bolts)

The available tearout strength for a pair of other bolts is:   0.75

LRFD

rn  0.75 143 kips   107 kips (for pair of bolts)

  2.00

ASD

rn 143 kips   2.00  71.5 kips (for pair of bolts)

Bolt shear strength controls over tearout and bearing strength for the other bolts in the angles.

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Strength of Bolted Connection The effective strength for the bolted connection at the angles is determined by summing the effective strength for each bolt using the minimum available strength calculated for bolt shear, bearing on the angles, and tearout on the angles. LRFD Rn  1 bolt  40.8 kips 

ASD Rn = 1 bolt  27.2 kips     4 bolts  32.5 kips 

  4 bolts  48.7 kips   236 kips  75 kips

o.k.

 157 kips  50 kips

o.k.

Shear and Tension Interaction in Bolts The required shear stress for each bolt is determined as follows:

f rv 

Vr nAb

where Ab  0.601 in.2 (from AISC Manual Table 7-1)

n  10 bolts LRFD

f rv 

ASD

75 kips

f rv 

10 bolts   0.601 in.2 

 12.5 ksi

50 kips

10 bolts   0.601 in.2 

 8.32 ksi

The nominal tensile stress modified to include the effects of shear stress is determined from AISC Specification Section J3.7 as follows. From AISC Specification Table J3.2: Fnt  90 ksi Fnv  54 ksi

LRFD

  0.75

  2.00

Fnt f rv  Fnt (Spec. Eq. J3-3a) Fnv 90 ksi  1.3  90 ksi   12.5 ksi   90 ksi 0.75  54 ksi 

Fnt  1.3Fnt 

 89.2 ksi  90 ksi

o.k.

ASD

Fnt f rv  Fnt (Spec. Eq. J3-3b) Fnv 2.00  90 ksi   1.3  90 ksi   8.32 ksi   90 ksi 54 ksi  89.3 ksi  90 ksi o.k.

Fnt  1.3Fnt 

Using the value of Fnt = 89.2 ksi determined for LRFD, the nominal tensile strength of one bolt is:

rn  Fnt Ab



  89.2 ksi  0.601 in.

2

(Spec. Eq. J3-2)



 53.6 kips

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The available tensile strength due to combined tension and shear is: LRFD

  0.75

  2.00

Rn r n n  

Rn  nrn  10 bolts  0.75  53.6 kips   402 kips  60 kips

ASD

 53.6 kips   10 bolts     2.00   268 kips  40 kips o.k.

o.k.

Prying Action on Bolts From AISC Manual Part 9, the available tensile strength of the bolts in the outstanding angle legs taking prying action into account is determined as follows: a 

angle leg  2  + tw  gage 2 4.00 in. 2    + 0.355 in.  52 in. 2

 1.43 in.

Note: If the distance from the bolt centerline to the edge of the supporting element is smaller than a = 1.43 in., use the smaller a in the following calculation. gage  t w  t 2 52 in.  0.355 in.  2 in.  2  2.32 in.

b

d   d   a    a  b    1.25b  b  2 2     d in. d in.  1.43 in.   1.25  2.32 in.  2 2  1.87 in.  3.34 in.  1.87 in.

(Manual Eq. 9-23)

d   b   b  b  2  

(Manual Eq. 9-18)

 2.32 in. 

d in. 2

 1.88 in.

b a 1.88 in.  1.87 in.  1.01

(Manual Eq. 9-22)



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Note that end distances of 14 in. are used on the angles, so p is the average pitch of the bolts: l n 142 in.  5  2.90 in.

p

Check: ps 2.90 in.  3 in. o.k. d   dh

 , in. d p , in.  1 2.90 in.  0.677

(Manual Eq. 9-20)

  1

The angle thickness required to develop the available strength of the bolt with no prying action as follows: LRFD Bc  40.2 kips/bolt (calculated previously)

ASD Bc  26.8 kips/bolt (calculated previously)

  0.90 

  1.67 

4 Bc b pFu

tc 

(Manual Eq. 9-26a)

4  40.2 kips/bolt 1.88 in.



tc  

0.90  2.90 in. 58 ksi 

 4 Bc b pFu

(Manual Eq. 9-26b)

1.67  4  26.8 kips/bolt 1.88 in.

 2.90 in. 58 ksi 

 1.41 in.

 1.41 in. 2  1   tc     1 (1  )  t    1.41 in.  2  1     1 0.677 1  1.01  2 in.    5.11

 

Because    1, the angles have insufficient strength to develop the bolt strength, therefore: 2

t  Q    1     tc  2

 2 in.    1  0.677   1.41 in.   0.211

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(Manual Eq. 9-28)

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The available tensile strength of the bolts, taking prying action into account is determined from AISC Manual Equation 9-27, as follows: LRFD rn  Bc Q   40.2 kips/bolt  0.211  8.48 kips/bolt

ASD rn  Bc Q    26.8 kips/bolt  0.211  5.65 kips/bolt

Rn  nrn  10 bolts  8.48 kips/bolt   84.8 kips  60 kips

o.k.

Rn r n n    10 bolts  5.65 kips/bolt   56.5 kips  40 kips

o.k .

Weld Design The resultant load angle on the weld is: LRFD 1 

N    tan  u   Vu   60 kips   tan 1    75 kips   38.7

ASD 1 

N    tan  a   Va   40 kips   tan 1    50 kips   38.7

From AISC Manual Table 8-8 for Angle = 30° (which will lead to a conservative result), using total beam setback of 2 in. + 4 in. = w in. (the 4 in. is included to account for mill underrun): l  142 in. kl  32 in. – w in.  2.75 in. kl l 2.75 in.  142 in.  0.190

k

x  0.027 by interpolation al  32 in.  xl  32 in. – 0.027 142 in.  3.11 in.

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al l 3.11 in.  142 in.  0.214

a

C  2.69 by interpolation

The required weld size is determined using AISC Manual Equation 8-21, as follows: LRFD Dmin

ASD

Ru  CC1l 

Dmin

96.0 kips 0.75  2.69 1142 in. 2 sides 

 1.64 sixteenths

 Ra  CC1l 

2.00  64.0 kips 

2.69 114 2 in. 2 sides 

 1.64 sixteenths

Use a x-in. fillet weld (minimum size from AISC Specification Table J2.4). Beam Web Strength at Fillet Weld The minimum beam web thickness required to match the shear rupture strength of a weld both sides to that of the base metal is: tmin  

6.19 Dmin Fu

(from Manual Eq. 9-3)

6.19 1.64 

65 ksi  0.156 in.  0.355 in.

o.k.

Shear Strength of Angles From AISC Specification Section J4.2(a), the available shear yielding strength of the angles is determined as follows: Agv   2 angles  lt   2 angles 142 in.2 in.  14.5 in.2

Rn  0.60Fy Agv



 0.60  36 ksi  14.5 in.

2

(Spec. Eq. J4-3)



 313 kips

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LRFD

  1.00

  1.50 

ASD



Rn  1.00  313 kips 

Rn 313 kips   1.50  209 kips  64.0 kips o.k.

 313 kips  96.0 kips o.k.

From AISC Specification Section J4.2(b), the available shear rupture strength of the angle is determined as follows. The effective net area is determined in accordance with AISC Specification Section B4.3b.

Anv   2 angles  l  n  dh  z in.  t   2 angles  142 in.  5 , in.  z in.  2 in.  9.50 in.2 Rn  0.60Fu Anv



 0.60  58 ksi  9.50 in.

2

(Spec. Eq. J4-4)



 331 kips LRFD

  0.75

  2.00

Rn  0.75  331 kips 

ASD

Rn 331 kips   2.00  166 kips  64.0 kips o.k.

 248 kips  96.0 kips o.k. Tensile Strength of Angles—Beam Web Side

From AISC Specification Section J4.1(a), the available tensile yielding strength of the angles is determined as follows: Ag   2 angles  lt   2 angles 142 in.2 in.  14.5in.2 Rn  Fy Ag

(Spec. Eq. J4-1)



  36 ksi  14.5 in.

2



 522 kips

  0.90 

LRFD

Rn  0.90  522 kips   470 kips  60 kips

o.k.

  1.67

Rn 522 kips   1.67  313 kips  40 kips

ASD

o.k.

From AISC Specification Sections J4.1(b), the available tensile rupture strength of the angles is determined as follows:

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Rn  Fu Ae

(Spec. Eq. J4-2)

Because the angle legs are welded to the beam web there is no bolt hole reduction and Ae = Ag; therefore, tensile rupture will not control. Block Shear Rupture Strength of Angles–Outstanding Legs The nominal strength for the limit state of block shear rupture of the angles assuming an L-shaped tearout relative to shear load, is determined as follows. The tearout pattern is shown in Figure II.A-2B-2.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant where leh  

2  angle leg   tw  gage 2 2  4 in. + 0.355 in.  52 in. 2

 1.43 in. Ant   2 angles  leh  0.5  d h  z in.   t    2 angles  1.43 in. – 0.5 , in.  z in.  2 in.  0.930 in.2

Agv   2 angles  lev   n  1 s   t    2 angles  14 in.   5  1 3 in.  2 in.  13.3 in.2

Fig. II.A-2B-2. Block shear rupture of outstanding legs of angles.

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(Spec. Eq. J4-5)

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Anv  Agv   2 angles  n  0.5 dh  z in. t   13.3 in.2 –  2 angles  5  0.5, in.  z in.2 in.  8.80 in.2 U bs  1.0

and















Rn  0.60  58 ksi  8.80 in.2  1.0  58 ksi  0.930 in.2  0.60  36 ksi  13.3 in.2  1.0  58 ksi  0.930 in.2



 360 kips  341 kips

Therefore: Rn  341 kips

The available block shear rupture strength of the angles is: LRFD

  0.75 

Rn  0.75  341 kips   256 kips  75 kips

ASD

  2.00 

Rn 341 kips  2.00   171 kips  50 kips

o.k.

o.k.

Shear Strength of Beam From AISC Specification Section J4.2(a), the available shear yield strength of the beam web is determined as follows: Agv  dtw  18.0 in. 0.355 in.  6.39 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  6.39 in.2



 192 kips

  1.00 

LRFD

  1.50 

Rn  1.00 192 kips   192 kips  75 kips

Rn 192 kips   1.50  128 kips  50 kips

o.k.

ASD

o.k.

The limit state of shear rupture of the beam web does not apply in this example because the beam is uncoped. Block Shear Rupture Strength of Beam Web

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Assuming a U-shaped tearout along the weld relative to the axial load, and a total beam setback of w in. (includes 4 in. tolerance to account for possible mill underrun), the nominal block shear rupture strength is determined as follows.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Ant  ltw

 142 in. 0.355 in.  5.15 in.2

Agv   2  32 in.  setback  tw   2  32 in.  w in. 0.355 in.  1.95 in.2 Because the angles are welded and there is no reduction for bolt holes:

Anv  Agv  1.95 in.2 Ubs = 1 and















Rn  0.60  65 ksi  1.95 in.2  1.0  65 ksi  5.15 in.2  0.60  50 ksi  1.95 in.2  1.0  65 ksi  5.15 in.2



 411 kips  393 kips

Therefore: Rn  393 kips

The available block shear rupture strength of the web is: LRFD

  0.75 

Rn  0.75  393 kips   295 kips  60 kips

o.k.

  2.00 

Rn 393 kips   2.00  197 kips  40 kips

ASD

o.k.

Tensile Strength of Beam From AISC Specification Section J4.1(a), the available tensile yielding strength of the beam is determined from AISC Specification Equation J4-1: Rn  Fy Ag

(Spec. Eq. J4-1)



  50 ksi  14.7 in.2



 735 kips

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The available tensile yielding strength of the beam is: LRFD

  0.90

  1.67 

Rn  0.90  735 kips   662 kips  60 kips

Rn 735 kips   1.67  440 kips  40 kips

o.k.

ASD

o.k.

From AISC Specification Section J4.1(b), determine the available tensile rupture strength of the beam. The effective net area is Ae = AnU, where U is determined from AISC Specification Table D3.1, Case 2. The value of x is determined by treating the W-shape as two channels back-to-back and finding the horizontal distance to the center of gravity of one of the channels from the centerline of the beam. (Note that the fillets are ignored.) x 



  Ax  A

 0.178 in. 18.0 in.  2  0.570 in.  

0.178 in.   7.50 in.   7.50 in. 2    2  0.570 in.    2 2  2    2  14.7 in.    2  

 1.13 in.

The connection length, l, used in the determination of U will be reduced by 4 in. to account for possible mill underrun. The shear lag factor, U, is: U  1  1

x l 1.13 in.

 3 in.  4 in.

 0.589

The minimum value of U can be determined from AISC Specification Section D3, where U is the ratio of the gross area of the connected element to the member gross area. U 

Ant Ag

 d  2t f  tw Ag

18.0 in.  2  0.570 in.   0.355 in.  14.7 in.2  0.407

AISC Specification Table D3.1, Case 2 controls, use U = 0.589. Because the angles are welded and there is no reduction for bolt holes: An  Ag  14.7 in.2

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Ae  AnU



2

 14.7 in.

(Spec. Eq. D3-1)

  0.589 

 8.66 in.2 Rn  Fu Ae



  65 ksi  8.66 in.2

(Spec. Eq. J4-2)



 563 kips

  0.75 

LRFD

Rn  0.75  563 kips   422 kips  60 kips

o.k.

  2.00 

Rn 563 kips   2.00  282 kips  40 kips

Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-3

ALL-WELDED DOUBLE-ANGLE CONNECTION

Given: Repeat Example II.A-1A using AISC Manual Table 10-3 and applicable provisions from the AISC Specification to verify the strength of an all-welded double-angle connection between an ASTM A992 W36231 beam and an ASTM A992 W1490 column flange, as shown in Figure II.A-3-1. Use 70-ksi electrodes and ASTM A36 angles.

Fig. II.A-3-1. Connection geometry for Example II.A-3. Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W36231

tw = 0.760 in. Column W1490 tf = 0.710 in. From ASCE/SEI 7, Chapter 2, the required strength is:

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LRFD Ru  1.2  37.5 kips   1.6 113 kips 

ASD Ra  37.5 kips  113 kips  151 kips

 226 kips Design of Weld between Beam Web and Angles

Use AISC Manual Table 10-3 (Welds A). Try x-in. weld size, l = 24 in. LRFD

ASD Rn  171 kips  151 kips o.k. 

Rn  257 kips  226 kips o.k. 

From AISC Manual Table 10-3, the minimum beam web thickness is:

tw min  0.286 in.  0.760 in. o.k. Design of Weld between Column Flange and Angles Use AISC Manual Table 10-3 (Welds B). Try 4-in. weld size, l = 24 in. LRFD

Rn  229 kips  226 kips o.k. 

ASD Rn  153 kips  151 kips o.k. 

From AISC Manual Table 10-3, the minimum column flange thickness is:

tf

min

 0.190 in.  0.710 in. o.k.

Angle Thickness Minimum angle thickness for weld from AISC Specification Section J2.2b: tmin  w  z in.  4 in.  z in.  c in.

Try 2L432c (SLBB). Shear Strength of Angles From AISC Specification Section J4.2(a), the available shear yielding strength of the angles is determined as follows: Agv   2 angles  lt   2 angles  24 in. c in.  15.0 in.2

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Rn  0.60 Fy Agv



 0.60  36 ksi  15.0 in.2

(Spec. Eq. J4-3)



 324 kips LRFD

  1.00 

 Rn  1.00  324 kips   324 kips  226 kips o.k.

 = 1.50



ASD

Rn 324 kips   1.50  216 kips  151 kips o.k.

From AISC Specification Section J4.2(b), the available shear rupture strength of the angles is determined as follows: Anv   2 angles  lt   2 angles  24 in. c in.  15.0 in.2 Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  58 ksi  15.0 in.2



 522 kips

  0.75 

LRFD

 Rn  0.75  522 kips   392 kips  226 kips o.k.

 = 2.00



ASD

Rn 522 kips  2.00   261 kips  151 kips o.k.

Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-4

ALL-BOLTED DOUBLE-ANGLE CONNECTION IN A COPED BEAM

Given: Use AISC Manual Table 10-1 to verify the available strength of an all-bolted double-angle connection between an ASTM A992 W1850 beam and an ASTM A992 W2162 girder web, as shown in Figure II.A-4-1, to support the following beam end reactions: RD = 10 kips RL = 30 kips The beam top flange is coped 2 in. deep by 4 in. long, lev = 14 in., leh = 1s in. Use ASTM A36 angles.

Fig. II.A-4-1. Connection geometry for Example II.A-4. Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam and girder ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-1 the geometric properties are as follows: Beam W1850

d = 18.0 in. tw = 0.355 in.

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Girder W2162 tw = 0.400 in. From AISC Specification Table J3.3, the hole diameter of a w-in.-diameter bolt in a standard hole is: dh = m in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2 10 kips   1.6  30 kips 

ASD

Ra  10 kips  30 kips  40.0 kips

 60.0 kips Connection Design

Tabulated values in AISC Manual Table 10-1 consider the limit states of bolt shear, bolt bearing and tearout on the angles, shear yielding of the angles, shear rupture of the angles, and block shear rupture of the angles. Try 3 rows of bolts and 2L5324 (SLBB). LRFD

ASD Rn  51.1 kips > 40.0 kips o.k. 

Rn  76.7 kips > 60.0 kips o.k.  Coped Beam Strength

From AISC Manual Part 9, the available coped beam web strength is the lesser of the limit states of flexural local web buckling, shear yielding, shear rupture, block shear rupture, and the sum of the effective strengths of the individual fasteners. From the Commentary to AISC Specification Section J3.6, the effective strength of an individual fastener is the lesser of the fastener shear strength, the bearing strength at the bolt holes and the tearout strength at the bolt holes. Flexural local web buckling of beam web As shown in AISC Manual Figure 9-2, the cope dimensions are: c = 4 in. dc = 2.00 in. e  c  setback  4 in.  2 in.  4.50 in. ho  d  d c  18.0 in.  2.00 in.  16.0 in.

c 4 in.  d 18.0 in.  0.222

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c 4 in.  ho 16.0 in.  0.250 Because

c  1.0 : d

c f  2  d   2  0.222 

(Manual Eq. 9-14a)

 0.444 Because

c  1.0 : ho 1.65

h  k  2.2  o   c 

(Manual Eq. 9-13a) 1.65

 16.0 in.   2.2    4 in.   21.7 ho tw 16.0 in.  0.355 in.  45.1



(Manual Eq. 9-11)

k1  fk  1.61

(Manual Eq. 9-10)

  0.444  21.7   1.61  9.63

 p  0.475  0.475

k1 E Fy

(Manual Eq. 9-12)

 9.63 29, 000 ksi  50 ksi

 35.5 2 p  2  35.5   71.0

Because p <  ≤ 2p, calculate the nominal flexural strength using AISC Manual Equation 9-7. The plastic section modulus of the coped section, Znet, is determined from Table IV-11 (included in Part IV of this document).

Z net  42.5 in.3

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M p  Fy Znet



  50 ksi  42.5 in.3



 2,130 kip-in. From AISC Manual Table 9-2:

Snet  23.4 in.3 M y  Fy Snet



  50 ksi  23.4 in.3



 1,170 kip-in.    M n  M p   M p  M y    1  p 

(Manual Eq. 9-7)

 45.1    2,130 kip-in.   2,130 kip-in.  1,170 kip-in.   1  35.5   1,870 kip-in.

Mn e 1,870 kip-in.  4.50 in.  416 kips

Rn 

LRFD

  0.90

Rn  0.90  416 kips   374 kips  60.0 kips

o.k.

  1.67

ASD

Rn 416 kips   1.67  249 kips  40.0 kips

o.k.

Shear Strength of Beam Web From AISC Specification Section J4.2(a), the available shear yielding strength of the beam web is determined as follows: Agv  ho tw  16.0 in. 0.355 in.  5.68 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  5.68 in.2



 170 kips

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LRFD

  1.00

  1.50

Rn  1.00 170 kips 

ASD

Rn 170 kips   1.50  113 kips  40.0 kips

 170 kips  60.0 kips o.k.

o.k.

From AISC Specification Section J4.2(b), the available shear rupture strength of the beam web is determined as follows: Anv   ho  3  d h + z in.  t w  16.0 in.  3 m in. + z in.   0.355 in.

 4.75 in.2

Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  65 ksi  4.75 in.2



 185 kips

  0.75

LRFD

  2.00

Rn  0.75 185 kips 

ASD

Rn 185 kips   2.00  92.5 kips  40.0 kips

 139 kips  60.0 kips o.k.

o.k.

Block Shear Rupture of Beam Web From AISC Specification Section J4.3, the block shear rupture strength of the beam web, assuming a total beam setback of w in. (includes 4 in. tolerance to account for possible mill underrun), is determined as follows.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Agv   lev  2 s  tw  14 in.  2  3.00 in.   0.355 in.  2.57 in.2 Anv  Agv  2.5  d h  z in. tw  2.57 in.2  2.5 m in.  z in. 0.355 in.  1.79 in.2 Ant  leh  4 in.(underrun)  0.5  d h  z in.  tw  1s in.  4 in.(underrun)  0.5 m  z in.   0.355 in.  0.333 in.2

The block shear reduction coefficient, Ubs, is 1.0 for a single row beam end connection as illustrated in AISC Specification Commentary Figure C-J4.2.

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













Rn  0.60  65 ksi  1.79 in.2  1.0  65 ksi  0.333 in.2  0.60  50 ksi  2.57 in.2  1.0  65 ksi  0.333 in.2



 91.5 kips  98.7 kips Therefore:

Rn  91.5 kips   0.75

LRFD

  2.00

Rn  0.75  91.5 kips 

ASD

Rn 91.5 kips   2.00  45.8 kips  40.0 kips

 68.6 kips  60.0 kips o.k.

o.k.

Strength of the Bolted Connection—Beam Web Side From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the invividual strengths of the individual fasteners, which may be taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear (or pair of bolts) is: LRFD

ASD rn  23.9 kips/bolt 

rn  35.8 kips/bolt

The available bearing and tearout strength of the beam web at Bolt 1, as shown in Figure II.A-4-1, is determine using AISC Manual Table 7-5 with le = 14 in. LRFD

rn   49.4 kip/in. 0.355 in.  17.5 kips/bolt

ASD rn   32.9 kip/in. 0.355 in.   11.7 kips/bolt

Therefore, bearing or tearout of the beam web controls over bolt shear for Bolt 1. The available bearing and tearout strength of the beam web at the other bolts is determine using AISC Manual Table 7-4 with s = 3 in. LRFD

rn   87.8 kip/in. 0.355 in.  31.2 kips/bolt

ASD rn   58.5 kip/in. 0.355 in.   20.8 kips/bolt

Therefore, bearing or tearout of the beam web controls over bolt shear for the other bolts. The strength of the bolt group in the beam web is determined by summing the strength of the individual fasteners as follows:

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LRFD

ASD

Rn

Rn  1 bolt 17.5 kips/bolt 



  2 bolts  31.2 kips/bolt   79.9 kips/bolt  60.0 kips o.k.

 1 bolt 11.7 kips/bolt    2 bolts  20.8 kips/bolt   53.3 kips/bolt  40.0 kips o.k.

Strength of the Bolted Connection—Support Side From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in single shear is: LRFD

ASD rn  11.9 kips/bolt 

rn  17.9 kips/bolt

Because the girder is not coped, the available bearing and tearout strength of the girder web at all bolts is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

rn   87.8 kip/in. 0.400 in.  35.1 kips/bolt

ASD rn   58.5 kip/in. 0.400 in.   23.4 kips/bolt

Therefore, bolt shear shear controls over bearing and tearout. Bolt shear strength is one of the limit states checked in previous calculations; thus, the effective strength of the fasteners is adequate. Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-5

WELDED/BOLTED DOUBLE-ANGLE CONNECTION IN A COPED BEAM

Given: Use AISC Manual Table 10-2 to verify the available strength of a double angle shear connection welded to an ASTM A992 W1850 beam and bolted to an ASTM A992 W2162 girder web, as shown in Figure II.A-5-1. Use 70-ksi electrodes and ASTM A36 angles.

Fig. II.A-5-1. Connection geometry for Example II.A-5. Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam and girder ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-1 the geometric properties are as follows: Beam W1850

d = 18.0 in. tw = 0.355 in. Girder W2162 tw = 0.400 in. From AISC Specification Table J3.3, the hole diameter of a w-in.-diameter bolt in a standard hole is:

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dh = m in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2 10 kips   1.6  30 kips 

ASD

Ra  10 kips  30 kips  40.0 kips

 60.0 kips Weld Design

Use AISC Manual Table 10-2 (Welds A). Try x-in. weld size, l = 82 in. LRFD

ASD Rn  73.5 kips  40.0 kips o.k. 

Rn  110 kips  60.0 kips o.k. 

From AISC Manual Table 10-2, the minimum beam web thickness is:

tw min  0.286 in.  0.355 in. o.k. Minimum Angle Thickness for Weld From AISC Specification Section J2.2b, the minimum angle thickness is: tmin  w  z in.  x in.  z in.  4 in.

Angle and Bolt Design Tabulated values in AISC Manual Table 10-1 consider the limit states of bolt shear, bolt bearing and tearout on the angles, shear yielding of the angles, shear rupture of the angles, and block shear rupture of the angles. Try 3 rows of bolts and 2L4324 (SLBB). LRFD

ASD Rn  51.1 kips > 40.0 kips o.k. 

Rn  76.7 kips  60.0 kips o.k.  Coped Beam Strength

The available flexural local web buckling strength of the coped beam is verified in Example II.A-4. Block Shear Rupture of Beam Web From AISC Specification Section J4.3, the block shear rupture strength of the beam web, assuming a total beam setback of w in. (includes 4 in. tolerance to account for possible mill underrun), is determined as follows.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

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(Spec. Eq. J4-5)

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where Agv   l  a in. tw   82 in.  a in. 0.355 in.  3.15 in.2

Anv  Agv  3.15 in.2 Ant   32 in.  w in. tw   32 in.  w in. 0.355 in.  0.976 in.2 U bs  1.0

and















Rn  0.60  65 ksi  3.15 in.2  1.0  65 ksi  0.976 in.2  0.60  50 ksi  3.15 in.2  1.0  65 ksi  0.976 in.2



 186 kips  158 kips Therefore:

Rn  158 kips LRFD

  0.75

Rn  0.75 158 kips   119 kips  60.0 kips o.k.

  2.00

ASD

Rn 158 kips   2.00  79.0 kips  40.0 kips

o.k.

Shear Strength of Beam Web From AISC Specification Section J4.2(a), the available shear yielding strength of the beam web is determined as follows: Agv   d  d c  tw  18.0 in.  2.00 in. 0.355 in.  5.68 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  5.68 in.2



 170 kips

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LRFD

  1.00

  1.50

Rn  1.00 170 kips 

ASD

Rn 170 kips   1.50  113 kips  40.0 kips

 170 kips  60.0 kips o.k.

o.k.

From AISC Specification Section J4.2(b), the available shear rupture strength of the beam web is determined as follows. Because the angle is welded to the beam web, there is no reduction for bolt holes, therefore: Anv  Agv  5.68 in.2

Rn  0.60 Fu Anv



 0.60  65 ksi  5.68 in.

2

(Spec. Eq. J4-4)



 222 kips   0.75

LRFD

  2.00

Rn  0.75  222 kips 

ASD

Rn 222 kips   2.00  111 kips  40.0 kips

 167 kips  60.0 kips o.k. Effective Strength of the Fasteners to the Girder Web

The effective strength of the fasteners to the girder web is verified in Example II.A-4. Summary The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-6

BEAM END COPED AT THE TOP FLANGE ONLY

Given: For an ASTM A992 W2162 beam coped 8 in. deep by 9 in. long at the top flange only, assuming a 2 in. setback (e = 9½ in.) and using an ASTM A572 Grade 50 plate for the stiffeners and doubler: A. Calculate the available strength of the beam end, as shown in Figure II.A-6-1(a), considering the limit states of flexural yielding, flexural local buckling, shear yielding and shear rupture. B. Choose an alternate ASTM A992 W21 shape to eliminate the need for stiffening for the following end reactions: RD = 23 kips RL = 67 kips C. Determine the size of doubler plate needed to reinforce the W2162, as shown in Figure II.A-6-1(c), for the given end reaction in Solution B. D. Determine the size of longitudinal stiffeners needed to stiffen the W21, as shown in Figure II.A-6-1(d), for the given end reaction in Solution B. Assume the shear connection is welded to the beam web.

Fig. II.A-6-1. Connection geometry for Example II.A-6. Solution A: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows:

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Beam W2162 ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A572 Grade 50 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1 the geometric properties are as follows: Beam W2162

d tw bf tf

= 21.0 in. = 0.400 in. = 8.24 in. = 0.615 in.

Coped Beam Strength The beam is assumed to be braced at the end of the uncoped section. Such bracing can be provided by a bracing member or by a slab or other suitable means. Flexural Local Buckling of Beam Web The limit state of flexural yielding and local web buckling of the coped beam web are checked using AISC Manual Part 9 as follows. ho  d  d c (from AISC Manual Figure 9-2)  21.0 in.  8.00 in.  13.0 in.

c 9.00 in.  d 21.0 in.  0.429

c 9.00 in.  ho 13.0 in.  0.692 Because

c  1.0, the buckling adjustment factor, f, is calculated as: d

c f  2  d   2  0.429 

(Manual Eq. 9-14a)

 0.858 Because

c  1.0, the plate buckling coefficient, k, is calculated as: ho

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1.65

h  k  2.2  o   c 

(Manual Eq. 9-13a) 1.65

 13.0 in.    9.00 in.   4.04

 2.2 

The modified plate buckling coefficient, k1, is calculated as:

k1  fk  1.61   0.858  4.04   1.61

(Manual Eq. 9-10)

 3.47

The plastic section modulus, Znet, is determined from Table IV-11 (included in Part IV of this document):

Z net  32.2 in.3 The plastic moment capacity, Mp, is: M p  Fy Z net



  50 ksi  32.2 in.3



 1, 610 kip-in.

The elastic section modulus, Snet, is determined from AISC Manual Table 9-2:

Snet  17.8 in.3 The flexural yield moment, My, is: M y  Fy S net



  50 ksi  17.8 in.3



 890 kip-in.

ho tw 13.0 in.  0.400 in.  32.5



 p  0.475  0.475

(Manual Eq. 9-11)

k1 E Fy

(Manual Eq. 9-12)

 3.47  29, 000 ksi  50 ksi

 21.3

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2 p  2  21.3   42.6

Because p <   2p, the nominal flexural strength is:    M n  M p   M p  M y    1  p 

(Manual Eq. 9-7)

 32.5   1, 610 kip-in.  1, 610 kip-in.  890 kip-in.   1  21.3   1, 230 kip-in.

The nominal strength of the coped section is: Mn e 1, 230 kip-in.  9.50 in.  129 kips

Rn 

The available strength of the coped section is: LRFD

  0.90

Rn  0.90 129 kips 

  1.67

ASD

Rn 129 kips   1.67  77.2 kips

 116 kips Shear Strength of Beam Web

From AISC Specification Section J4.2(a), the available shear yielding strength of the beam web is determined as follows: Agv   d  d c  tw   21.0 in.  8.00 in. 0.400 in.  5.20 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  5.20 in.

2



 156 kips

  1.00

Rn  1.00 156 kips   156 kips

LRFD

  1.50

Rn 156 kips  1.50   104 kips

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From AISC Specification Section J4.2(b), the available shear rupture strength of the beam web is determined as follows. Because the connection is welded to the beam web there is no reduction for bolt holes, therefore: Anv  Agv

 5.20 in.2

Rn  0.60 Fu Anv



 0.60  65 ksi  5.20 in.

2

(Spec. Eq. J4-4)



 203 kips   0.75

LRFD

  2.00

ASD

Rn 203 kips   2.00  102 kips

Rn  0.75  203 kips   152 kips

Thus, the available strength of the beam is controlled by the coped section. LRFD

ASD Rn  77.2 kips 

Rn  116 kips  Solution B:

From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  23 kips   1.6  67 kips   135 kips

ASD

Ra  23 kips  67 kips  90.0 kips

Try a W2173. From AISC Manual Table 2-4, the material properties are as follows: Beam W2173

ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1 the geometric properties are as follows: Beam W2173 d = 21.2 in. tw = 0.455 in. bf = 8.30 in. tf = 0.740 in. Flexural Local Buckling of Beam Web

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The limit state of flexural yielding and local web buckling of the coped beam web are checked using AISC Manual Part 9 as follows. ho  d  d c (from AISC Manual Figure 9-2)  21.2 in.  8.00 in.  13.2 in.

c 9.00 in.  d 21.2 in.  0.425

c 9.00 in.  ho 13.2 in.  0.682 Because

c  1.0, the buckling adjustment factor, f, is calculated as: d

c f  2  d   2  0.425 

(Manual Eq. 9-14a)

 0.850 Because

c  1.0, the plate buckling coefficient, k, is calculated as: ho 1.65

h  k  2.2  o   c 

(Manual Eq. 9-13a) 1.65

 13.2 in.   2.2    9.00 in.   4.14

The modified plate buckling coefficient, k1, is calculated as:

k1  fk  1.61

(Manual Eq. 9-10)

  0.850  4.14   1.61  3.52 The plastic section modulus, Znet, is determined from Table IV-11 (included in Part IV of this document):

Z net  37.6 in.3 The plastic moment capacity, Mp, is: M p  Fy Z net



  50 ksi  37.6 in.3



 1,880 kip-in.

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The elastic section modulus, Snet, is determined from AISC Manual Table 9-2:

Snet  21.0 in.3 The flexural yield moment, My, is: M y  Fy S net



  50 ksi  21.0 in.3



 1, 050 kip-in.

ho tw 13.2 in.  0.455 in.  29.0



 p  0.475  0.475

(Manual Eq. 9-11)

k1 E Fy

(Manual Eq. 9-11)

 3.52  29, 000 ksi  50 ksi

 21.5 2 p  2  21.5   43.0

Since p <   2p, the nominal flexural strength is:     1 M n  M p   M p  M y    p 

(Manual Eq. 9-7)

 29.0   1,880 kip-in.  1,880 kip-in.  1, 050 kip-in.   1  21.5   1,590 kip-in.

The nominal strength of the coped section is: Mn e 1,590 kip-in.  9.50 in.  167 kips

Rn 

The available strength of the coped section is:

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LRFD

  0.90

Rn  0.90 167 kips 

  1.67

ASD

Rn 167 kips   1.67  100 kips

 150 kips Shear Strength of Beam Web

From AISC Specification Section J4.2(a), the available shear yielding strength of the beam web is determined as follows: Agv   d  d c  tw   21.2 in.  8.00 in. 0.455 in.  6.01 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  6.01 in.

2



 180 kips

LRFD

  1.00

  1.50

ASD

Rn 180 kips   1.50  120 kips

Rn  1.00 180 kips   180 kips

From AISC Specification Section J4.2(b), the available shear rupture strength of the beam web is determined as follows. Because the connection is welded to the beam web, there is no reduction for bolt holes, therefore: Anv  Agv

 6.01 in.2 Rn  0.60 Fu Anv



 0.60  65 ksi  6.01 in.

2

(Spec. Eq. J4-4)



 234 kips

  0.75

Rn  0.75  234 kips   176 kips

LRFD

  2.00

ASD

Rn 234 kips   2.00  117 kips

Thus, the available strength is controlled by the coped section, therefore the available strength of the beam is:

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LRFD

ASD Rn  100 kips  90.0 kips o.k. 

Rn  150 kips  135kips o.k.  Solution C: Doubler Plate Design

The doubler plate is designed using AISC Manual Part 9. An ASTM A572 Grade 50 plate is recommended in order to match the beam yield strength. A 4-in. minimum plate thickness will be used in order to allow the use of a x-in. fillet weld. The depth of the plate will be set so that a compact b/t ratio from AISC Specification Table B4.1b will be satisfied. This is a conservative criterion that will allow local buckling of the doubler to be neglected. dp E  1.12 tp Fy

Solving for dp: d p  1.12t p

E Fy

 1.12  0.250 in.

29, 000 ksi 50 ksi

 6.74 in.

A 6.50 in. doubler plate will be used. Using principles of mechanics, the elastic section modulus, Snet, and plastic section modulus, Znet, are calculated neglecting the fillets and assuming the doubler plate is placed 2-in. down from the top of the cope. S net  25.5 in.3 Z net  44.8 in.3

The plastic bending moment, Mp, of the reinforced section is: M p  Fy Z net



  50 ksi  44.8 in.3



 2, 240 kip-in.

The flexural yield moment, My, of the reinforced section is: M y  Fy S net



  50 ksi  25.5 in.3



 1, 280 kip-in.

Because p <   2p for the unreinforced section, the nominal flexural strength is:

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    1 M n  M p   M p  M y    p 

(Manual Eq. 9-7)

 32.5   2, 240 kip-in.   2, 240 kip-in.  1, 280 kip-in.   1  21.3   1, 740 kip-in.

The available strength of the coped section is determined as follows: Mn e 1, 740 kip-in.  9.50 in.  183 kips

Rn 

LRFD

  0.90

  1.67

Rn  0.90 183 kips 

ASD

Rn 183 kips   1.67  110 kips

 165 kips Shear Strength of Beam Web

From AISC Specification Section J4.2(a), the available shear yielding strength of the beam web reinforced with the doubler plate is determined as follows: Agv  web   d  dc  tw   21.0 in.  8.00 in. 0.400 in.  5.20 in.2

Agv  plate  d p t p

  6.50 in.4 in.  1.63 in.2 Rn  0.60 Fy Agv  web  0.60 Fy Agv  plate





(from Spec. Eq. J4-3)



 0.60  50 ksi  5.20 in.2  0.60  50 ksi  1.63 in.2



 205 kips

  1.00

Rn  1.00  205 kips   205 kips

LRFD

  1.50

ASD

Rn 205 kips   1.50  137 kips

From AISC Specification Section J4.2(b), the available shear rupture strength of the beam web reinforced with the doubler plate is determined as follows. Because the connection is welded, there is no reduction for bolt holes, therefore:

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Anv  web  Agv  web

 5.20 in.2 Anv  plate  Agv  plate

 1.63 in.2 Rn  0.60 Fu Anv  web  0.60 Fu Anv  plate





(from Spec. Eq. J4-4)



 0.60  65 ksi  5.20 in.2  0.60  65 ksi  1.63 in.2



 266 kips

  0.75

LRFD

ASD

  2.00

Rn  0.75  266 kips 

Rn 266 kips   2.00  133 kips

 200 kips

Thus, the available strength of the beam is controlled by the coped section. LRFD

ASD Rn  110 kips  90.0 kips o.k. 

Rn  165 kips  135kips o.k.  Weld Design

Determine the length of weld required to transfer the force into and out of the doubler plate. From Solution A, the available strength of the beam web is: LRFD

ASD Rn  77.2 kips 

Rn  116 kips

The available strength of the beam web reinforced with the doubler plate is: LRFD

ASD Rn  110 kips 

Rn  165 kips The force in the doubler plate is determined as follows:   0.90

LRFD

ASD

  1.67

 116 kips  Fd  0.90  50 ksi 4 in. 6.50 in.    165 kips   51.4 kips

 77.2 kips    110 kips 

 50 ksi 4 in. 6.50 in.  Fd 

1.67

 34.1 kips

From AISC Specification Section J2.4, the doubler plate weld is designed as follows:

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Rn  0.85 Rnwl  1.5 Rnwt

(Spec. Eq. J2-6b)

LRFD From AISC Manual Equation 8-2a:

ASD From AISC Manual Equation 8-2b:

Rnw  1.392 Dl

Rnw  0.928 Dl

From AISC Specification Equation J2-6b:

From AISC Specification Equation J2-6b:

 2 welds  0.851.392 kips/in.  51.4 kips      3 sixteenths  lw  1.51.392 kips/in. 3 sixteenths       6.50 in. 

 2 welds  0.85 0.928 kips/in.  34.1 kips      3 sixteenths  lw  1.5  0.928 kips/in. 3 sixteenths       6.50 in. 

Solving for lw:

Solving for lw:

lw = 1.50 in.

lw = 1.47 in..

Use 1.50 in. of x-in. fillet weld, minimum. The doubler plate must extend at least dc beyond the cope. Use a PL4 in. 62 in. 1ft 5 in. with x-in. welds all around. Solution D: Longitudinal Stiffener Design

Try PL4 in.4 in. slotted to fit over the beam web. Determine Zx for the stiffened section: Aw   d  d c  t f  tw   21.0 in.  8.00 in.  0.615 in. 0.400 in.  4.95 in.2

Af  b f t f

  8.24 in. 0.615 in.  5.07 in.2 Arp  b p t p

  4.00 in.4 in.  1.00 in.2

At  Aw  A f  Arp  4.95 in.2  5.07 in.2  1.00 in.2  11.0 in.2

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The location of the plastic neutral axis (neglecting fillets) from the inside of the flange is:

 0.615 in.8.24 in.  y p  0.400 in.  4 in. 4.00 in.  12.4 in.  y p   0.400 in. y p  1.12 in. From elementary mechanics, the section properties are as follows: Zx = 44.3 in.3 Ix = 253 in.4 Sxc = 28.6 in.3 Sxt = 57.7 in.3

hc  2 13.0 in.  4.39 in.  17.2 in. hp  2 13.0 in.  1.12 in.  0.615 in.  22.5 in. Compact section properties for the longitudinal stiffener and the web are determined from AISC Specification Table B4.1b, Cases 11 and 16.  p  0.38  0.38

E Fy

(Spec. Table B4.1b, Case 11)

29, 000 ksi 50 ksi

 9.15

 

b t  4.00 in. 2 

4 in.  8.00 Because    p , the stiffener is compact in flexure.  r  5.70  5.70

E Fy

(Spec. Table B4.1b, Case 16)

29, 000 ksi 50 ksi

 137

hc tw 17.2 in.  0.400 in.  43.0



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Because    r , the web is not slender, therefore AISC Specification Section F4 applies. Determine if lateral-torsional buckling is a design consideration. aw  

hc tw b fc t fc

(Spec. Eq. F4-12)

17.2 in. 0.400 in.  4.00 in.4 in.

 6.88

b fc

rt 

(Spec. Eq. F4-11)

 1  12 1  aw   6  4.00 in.



 1  12 1   6.88   6   0.788 in. L p  1.1rt

E Fy

(Spec. Eq. F4-7)

 1.1 0.788 in.

29, 000 ksi 50 ksi

 20.9 in.

The stiffener will not reach a length of 20.9 in. Lateral-torsional buckling is not a design consideration. Determine if the web of the singly-symmetric shape is compact. AISC Specification Table B4.1b, Case 16, applies.

p 





hc hp

E Fy

  Mp  0.09   0.54 M y  

2

 5.70

E Fy

17.2 in. 29, 000 ksi 22.5 in. 50 ksi

   2, 220 kip-in.  0.54    0.09   1, 430 kip-in.     32.9  137  32.9



2

 5.70

29, 000 ksi 50 ksi

hc tw 17.2 in.  0.400 in.  43.0

 

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Because    p , the web is non-compact, therefore AISC Specification Section F4 applies. Since Sxt > Sxc, tension flange yielding does not govern. Determine flexural strength based on compression flange yielding. M yc  S xc Fy





 28.6 in.3  50 ksi   1, 430 kip-in.

I yc 

4 in. 4.00 in.3 12 4

 1.33 in.

I y  1.33 in.4 

 0.615 in.8.24 in.3 12.4 in. 0.400 in.3 12



12

4

 30.1 in. I yc Iy

Since



1.33 in.4

30.1 in.4  0.0442

I yc < 0.23, Rpc = 1.0. Thus: Iy

M n  R pc M yc  1.0 1, 430 kip-in.  1, 430 kip-in. The nominal strength of the reinforced section is: Mn e 1, 430 kip-in.  9.50 in.  151 kips

Rn 

  0.90

LRFD

Rn  0.90 151 kips   136 kips  135 kips o.k.

  1.67

ASD

Rn 151 kips   1.67  90.4 kips  90.0 kips

o.k.

Plate Dimensions Since the longitudinal stiffening must extend at least dc beyond the cope, use PL4 in.4 in.1 ft 5 in. with 4-in. welds.

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Weld Strength By calculations not shown, the moment of inertia of the reinforced section and distance from the centroid to the bottom of the reinforcement plate are:

I net  253 in.4 y  8.61 in.

The first moment of the reinforcement plate is: Q  Ap y  4 in. 4.00 in. 8.61 in.  0.5 4 in.   8.74 in.3

where Ap is the area of the reinforcement plate and y is the distance from the centroid of the reinforced section to the centroid of the reinforcement plate. From mechanics of materials and shear flow, the force per length that the weld must resist in the area of the cope is: LRFD

ASD

Vu Q ru  I net  2 welds 

Va Q ra  I net  2 welds 

 2.33 kip/in.

 1.55 kip/in.

135 kips   8.74 in.3    253 in.4   2 welds 

 90.0 kips  8.74 in.3    253 in.4   2 welds 

From mechanics of materials, the force per length that the weld must resist to transfer the force in the reinforcement plate to the beam web is: LRFD Vu eQ ru  I net  2 welds  l  c 

ASD

135 kips  9.50 in. 8.74 in.3    253 in.4   2 welds 17.0 in.  9.00 in.  2.77 kip/in.

Va eQ ra  I net  2 welds  l  c 

 90.0 kips  9.50 in.  8.74 in.3    253 in.4   2 welds 17.0 in.  9.00 in.  1.85 kip/in. controls

controls

The weld capacity from AISC Manual Part 8:

rn  1.392 kip/in. D

LRFD

ASD (from Manual Eq. 8-2a)

 1.392 kip/in. 4 sixteenths   5.57 kip/in.  2.77 kip/in.

o.k.

rn   0.928 kip/in. D (from Manual Eq. 8-2b)    0.928 kip/in. 4 sixteenths 

 3.71 kip/in.  1.85 kip/in.

Determine if the web has adequate shear rupture capacity:

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  0.75

LRFD

rn  0.60 Fu Anv =

  2.00

(from Spec. Eq. J4-4)

0.75  0.60  65 ksi  0.400 in.

2 welds  5.85 kip/in.  2.77 kip/in.

o.k.

ASD

rn 0.60 Fu Anv    0.60  65 ksi  0.400 in. = 2.00  2 welds   5.85 kip/in.  1.85 kip/in.

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(from Spec. Eq. J4-4)

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EXAMPLE II.A-7

BEAM END COPED AT THE TOP AND BOTTOM FLANGES

Given:

Determine the available strength for an ASTM A992 W1640 coped 32 in. deep by 92 in. wide at the top flange and 2 in. deep by 92 in. wide at the bottom flange, as shown in Figure II.A-7-1, considering the limit states of flexural yielding and local buckling. Assume a 2-in. setback from the face of the support to the end of the beam.

Fig. II.A-7-1. Connection geometry for Example II.A-7. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beam W1640

ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1 and AISC Manual Figure 9-3, the geometric properties are as follows: Beam W1640

d = 16.0 in. tw = 0.305 in. tf = 0.505 in. bf = 7.00 in. ct = 92in. dct = 32 in. cb = 92 in. dcb = 2 in. e = 92 in. + 2 in. = 10.0 in. ho = d – dct – dcb = 16.0 in. - 32 in. – 2 in. = 10.5 in.

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For a beam that is coped at both flanges, the local flexural strength is determined in accordance with AISC Specification Section F11. Available Strength at Coped Section

The cope at the tension side of the beam is equal to the cope length at the compression side. From AISC Manual Part 9, Lb = ct and dct is the depth of the cope at the top flange.

  L   d  Cb  3  ln  b   1  ct   1.84 d   d      92 in.    32 in.   3  ln    1    1.84  16.0 in.    16.0 in.    1.94  1.84

(Manual Eq. 9-15)

Use Cb = 1.84. The available strength of the coped section is determined using AISC Specification Section F11, with d = ho = 10.5 in. and unbraced length Lb = ct = 92 in. Lb d t

2



 92 in.10.5 in.  0.305 in.2

 1, 070

0.08 E 0.08  29, 000 ksi   50 ksi Fy  46.4

1.9 E 1.9  29, 000 ksi   Fy 50 ksi  1,100 0.08 E Lb d 1.9 E  2  , the limit state of lateral-torsional buckling applies. The nominal flexural strength of Fy Fy t the coped portion of the web is determined using AISC Specification Section F11.2(b).

Since

Determine the net elastic and plastic section moduli: S net  

tw ho 2 6

 0.305 in.10.5 in.2 6 3

 5.60 in.

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Z net  

tw ho 2 4

 0.305 in.10.5 in.2 4 3

 8.41 in. M y  Fy S net



  50 ksi  5.60 in.3



 280 kip-in. M p  Fy Z net



  50 ksi  8.41 in.3



 421 kip-in.

  L d  Fy  M n  Cb 1.52  0.274  b2   M y  M p  t  E    50 ksi    1.84 1.52  0.274 1, 070      280 kip-in.  421 kip-in.  29, 000 ksi   

(Spec. Eq. F11-2)

 523 kip-in.  421 kip-in.

The nominal moment capacity of the reduced section is 421 kip-in. The nominal strength of the coped section is: Mn e 421 kip-in.  10.0 in.  42.1 kips

Rn 

The available strength at the coped end is: LRFD

ASD

b  0.90

b  1.67

b Rn  0.90  42.1 kips 

Rn 42.1 kips  b 1.67  25.2 kips

 37.9 kips

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EXAMPLE II.A-8

ALL-BOLTED DOUBLE-ANGLE CONNECTIONS (BEAMS-TO-GIRDER WEB)

Given: Verify the all-bolted double-angle connections for back-to-back ASTM A992 W1240 and W2150 beams to an ASTM A992 W3099 girder-web to support the end reactions shown in Figure II.A-8-1. Use ASTM A36 angles.

Fig. II.A-8-1. Connection geometry for Example II.A-8. Solution: From AISC Manual Table 2-4, the material properties are as follows: Beams and girder ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1 the geometric properties are as follows:

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Beam W1240 tw = 0.295 in. d = 11.9 in. Beam W2150 tw = 0.380 in. d = 20.8 in. Girder W3099 tw = 0.520 in. d = 29.7 in. From AISC Specification Table J3.3, for w-in.-diameter bolts with standard holes: dh = m in. Beam A Connection: From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  4.17 kips   1.6 12.5 kips 

ASD Ra  4.17 kips  12.5 kips  16.7 kips

 25.0 kips Strength of Bolted Connection—Angles

AISC Manual Table 10-1 includes checks for the limit states of bolt shear, bolt bearing on the angles, tearout on the angles, shear yielding of the angles, shear rupture of the angles, and block shear rupture of the angles. For two rows of bolts and 4-in. angle thickness: LRFD Rn  48.9 kips  25.0 kips

ASD Rn  32.6 kips  16.7 kips o.k. 

o.k. 

Strength of the Bolted Connection—Beam Web From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the invividual strengths of the individual fasteners, taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear is: LRFD

rn  35.8 kips/bolt

ASD rn  23.9 kips/bolt 

The available bearing and tearout strength of the beam web at the top bolt is determined using AISC Manual Table 7-5, with le = 2 in., as follows:

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LRFD

ASD rn   58.5 kip/in. 0.295 in.   17.3 kips/bolt

rn   87.8 kip/in. 0.295 in.  25.9 kips/bolt

The available bearing and tearout strength of the beam web at the bottom bolt (not adjacent to the edge) is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

ASD rn   58.5 kip/in. 0.295 in.   17.3 kips/bolt

rn   87.8 kip/in. 0.295 in.  25.9 kips/bolt

The bearing or tearout strength controls over bolt shear for both bolts in the beam web. The strength of the bolt group in the beam web is determined by summing the strength of the individual fasteners as follows: LRFD

ASD Rn  1 bolt 17.3 kips/bolt    1 bolt 17.3 kips/bolt 

Rn  1 bolt  25.9 kips/bolt   1 bolt  25.9 kips/bolt   51.8 kips  25.0 kips o.k.

 34.6 kips  16.7 kips

o.k.

Coped Beam Strength From AISC Manual Part 9, the available coped beam web strength is the lesser of the limit states of flexural local web buckling, shear yielding, shear rupture, and block shear rupture. Flexural local web buckling of beam web The limit state of flexural yielding and local web buckling of the coped beam web are checked using AISC Manual Part 9 as follows: e  c  setback  5 in.  2 in.  5.50 in.

ho  d  d c (from AISC Manual Figure 9-2)  11.9 in.  2 in.  9.90 in.

c 5 in.  d 11.9 in.  0.420 c 5 in.  ho 9.90 in.  0.505

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Because

c  1.0, the buckling adjustment factor, f, is calculated as follows: d

c f  2  d   2  0.420 

(Manual Eq. 9-14a)

 0.840

Because

c  1.0, the plate buckling coefficient, k, is calculated as follows: ho 1.65

h  k  2.2  o   c 

(Manual Eq. 9-13a) 1.65

 9.90 in.   2.2    5 in.   6.79

ho tw 9.90 in.  0.295 in.  33.6



(Manual Eq. 9-11)

k1  fk  1.61

(Manual Eq. 9-10)

  0.840  6.79   1.61  5.70  1.61

 p  0.475  0.475

k1 E Fy

(Manual Eq. 9-12)

 5.70  29, 000 ksi  50 ksi

 27.3

2 p  2  27.3  54.6 Because p <  ≤ 2p, calculate the nominal moment strength using AISC Manual Equation 9-7. The plastic section modulus of the coped section, Znet, is determined from Table IV-11 (included in Part IV of this document).

Z net  14.0 in.3

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M p  Fy Z net



  50 ksi  14.0 in.3



 700 kip-in. From AISC Manual Table 9-2:

Snet  8.03 in.3

M y  Fy Snet



  50 ksi  8.03 in.3



 402 kip-in.    M n  M p   M p  M y    1   p 

(Manual Eq. 9-7)

 33.6    700 kip-in.   700 kip-in.  402 kip-in.    1  27.3    631 kip-in.

Mn e 631 kip-in.  5.50 in.

Rn 

 115 kips The available strength of the coped section is: LRFD

  0.90

Rn  0.90 115 kips   104 kips  25.0 kips

o.k.

  1.67

ASD

Rn 115 kips   1.67  68.9 kips  16.7 kips o.k.

Shear strength of beam web From AISC Specification Section J4.2, the available shear yielding strength of the beam web is determined as follows:

Agv  ho tw   9.90 in. 0.295 in.  2.92 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  2.92 in.2



 87.6 kips

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LRFD

  1.00

  1.50

Rn  1.00  87.6 kips 

ASD

Rn 87.6 kips   1.50  58.4 kips  16.7 kips o.k.

 87.6 kips  25.0 kips o.k.

From AISC Specification Section J4.2, the available shear rupture strength of the beam web is determined as follows: Anv   ho  n  d h + z in.  t w   9.90 in.  2 m in. + z in.   0.295 in.  2.40 in.2

Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  65 ksi  2.40 in.

2



 93.6 kips

  0.75

LRFD

  2.00

Rn  0.75  93.6 kips 

ASD

Rn 93.6 kips   2.00  46.8 kips  16.7 kips o.k.

 70.2 kips  25.0 kips o.k. Block shear rupture of beam web

The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the beam web is determined as follows, using AISC Manual Tables 93a, 9-3b and 9-3c and AISC Specification Equation J4-5, with n = 2, leh = 1a in. (includes 4-in. tolerance to account for possible beam underrun), lev = 2 in. and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a: 

Fu Ant  45.7 kip/in.  t





Shear yielding component from AISC Manual Table 9-3b: 

0.60Fy Agv  113 kip/in.   t

ASD Tension rupture component from AISC Manual Table 9-3a:

Fu Ant  30.5 kip/in. t

Shear yielding component from AISC Manual Table 9-3b:





0.60Fy Agv  75.0 kip/in.  t

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LRFD Shear rupture component from AISC Manual Table 9-3c:



0.60Fu Anv  108 kip/in.   t  The design block shear rupture strength is:

ASD Shear rupture component from AISC Manual Table 9-3c:





The allowable block shear rupture strength is: Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant  +     71.9 kip/in.  30.5 kip/in. 0.295 in.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant



0.60Fu Anv  71.9 kip/in. t

 108 kip/in.  45.7 kip/in. 0.295 in.  113 kip/in.  45.7 kip/in. 0.295 in.  45.3 kips  46.8 kips

  75.0 kip/in.  30.5 kip/in. 0.295 in.



 30.2 kips  31.1 kips

 Therefore:

Therefore: Rn  45.3 kips  25.0 kips

o.k.

Rn  30.2 kips  16.7 kips o.k. 

Beam B Connection:

From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2 18.3 kips   1.6  55 kips 

ASD

Ra  18.3 kips  55 kips  73.3 kips

 110 kips Strength of the Bolted Connection—Angles

AISC Manual Table 10-1 includes checks for the limit states of bolt shear, bolt bearing on the angles, tearout on the angles, shear yielding of the angles, shear rupture of the angles, and block shear rupture of the angles. For five rows of bolts and 4-in. angle thickness: LRFD Rn  126 kips  110 kips

ASD Rn  83.8 kips  73.3 kips o.k. 

o.k. 

Strength of the Bolted Connection—Beam Web From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear is: LRFD

rn  35.8 kips/bolt

ASD rn  23.9 kips/bolt 

The available bearing and tearout strength of the beam web at the top edge bolt is determined using AISC Manual Table 7-5 with le = 2 in.

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LRFD

ASD rn   58.5 kip/in. 0.380 in.   22.2 kips/bolt

rn   87.8 kip/in. 0.380 in.  33.4 kips/bolt

The available bearing and tearout strength of the beam web at the interior bolts (not adjacent to the edge) is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

ASD rn   58.5 kip/in. 0.380 in.   22.2 kips/bolt

rn   87.8 kip/in. 0.380 in.  33.4 kips/bolt

The strength of the bolt group in the beam web is determined as follows: LRFD

ASD Rn  1 bolt  22.2 kips/bolt     4 bolt  22.2 kips/bolt 

R  1 bolt  33.4 kips/bolt    4 bolts  33.4 kips/bolt   167 kips  110 kips o.k.

 111 kips  73.3 kips

o.k.

Coped Beam Strength From AISC Manual Part 9, the available coped beam web strength is the lesser of the limit states of flexural local web buckling, shear yielding, shear rupture, and block shear rupture. Flexural local web buckling of beam web The limit state of flexural yielding and local web buckling of the coped beam web are checked using AISC Manual Part 9 as follows: e  c  setback  5 in.  2 in.  5.50 in.

ho  d  d c (from AISC Manual Figure 9-2)  20.8 in.  2 in.  18.8 in.

c 5 in.  d 20.8 in.  0.240 c 5 in.  ho 18.8 in.  0.266

Because

c  1.0, the buckling adjustment factor, f, is calculated as follows: d

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c f  2  d   2  0.240 

(Manual Eq. 9-14a)

 0.480

Because

c  1.0, the plate buckling coefficient, k, is calculated as follows: ho 1.65

h  k  2.2  o   c 

(Manual Eq. 9-13a) 1.65

 18.8 in.   2.2    5 in.   19.6

ho tw 18.8 in.  0.380 in.  49.5



(Manual Eq. 9-11)

k1  fk  1.61

(Manual Eq. 9-10)

  0.480 19.6   1.61  9.41  1.61

 p  0.475  0.475

k1 E Fy

(Manual Eq. 9-12)

 9.41 29, 000 ksi  50 ksi

 35.1

2 p  2  35.1  70.2 Because p <  ≤ 2p, calculate the nominal moment strength using AISC Manual Equation 9-7. The plastic section modulus of the coped section, Znet, is determined from Table IV-11 (included in Part IV of this document).

Z net  56.5 in.3

M p  Fy Z net



  50 ksi  56.5 in.3



 2,830 kip-in. From AISC Manual Table 9-2:

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Snet  32.5 in.3

M y  Fy Snet



  50 ksi  32.5 in.3



 1, 630 kip-in.    M n  M p   M p  M y    1  p 

(Manual Eq. 9-7)

 49.5    2,830 kip-in.   2,830 kip-in.  1, 630 kip-in.    1  35.1    2, 340 kip-in.

Mn e 2,340 kip-in.  5.50 in.

Rn 

 425 kips LRFD

  0.90

Rn  0.90  425 kips   383 kips  110 kips

o.k.

  1.67

ASD

Rn 425 kips   1.67  254 kips  73.3 kips o.k.

Shear strength of beam web From AISC Specification Section J4.2, the available shear yielding strength of the beam web is determined as follows: Agv  ho tw  18.8 in. 0.380 in.  7.14 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  7.14 in.2



 214 kips

  1.00

LRFD

Rn  1.00  214 kips   214 kips  110 kips

o.k.

  1.50

ASD

Rn 214 kips   1.50  143 kips  73.3 kips o.k.

From AISC Specification Section J4.2, the available shear rupture strength of the beam web is determined as follows:

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Anv   ho  n  d h + z in.  t w  18.8 in.  5 m in. + z in.   0.380 in.  5.48 in.2

Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  65 ksi  5.48 in.

2



 214 kips

  0.75

LRFD

  2.00

Rn  0.75  214 kips 

ASD

Rn 214 kips   2.00  107 kips  73.3 kips o.k.

 161 kips  110 kips o.k. Block shear rupture of beam web

The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the beam web is determined as follows, using AISC Manual Tables 9-3a, 9-3b and 9-3c and AISC Specification Equation J4-5, with n = 5, leh = 1a in. (includes 4 in. tolerance to account for possible beam underrun), lev = 2 in. and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a: 



Fu Ant  45.7 kip/in. t

ASD Tension rupture component from AISC Manual Table 9-3a:





Fu Ant  30.5 kip/in. t

Shear yielding component from AISC Manual Table 9-3b: 

Shear yielding component from AISC Manual Table 9-3b:





0.60Fy Agv  315 kip/in.  t

0.60Fy Agv  210 kip/in.  t





Shear rupture component from AISC Manual Table 9-3c:

Shear rupture component from AISC Manual Table 9-3c:



0.60Fu Anv  294 kip/in. t



0.60Fu Anv  196 kip/in. t

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LRFD Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant   294 kip/in.  45.7 kip/in. 0.380 in.   315 kip/in.  45.7 kip/in. 0.380 in.  129 kips  137 kips

ASD Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant +     196 kip/in.  30.5 kip/in. 0.380 in.   210 kip/in.  30.5 kip/in. 0.380 in.  86.1 kips  91.4 kips

Therefore:

Therefore:

Rn  129 kips  110 kips

o.k.

Rn  86.1 kips  73.3 kips 

o.k.

Supporting Girder Connection

Supporting Girder Web The required effective strength per bolt is the minimum from the limit states of bolt shear, bolt bearing and tearout. The bolts that are loaded by both connections will have the largest demand.. Thus, for the design of these four critical bolts, the required strength is determined as follows: LRFD From the W1240 beam, each bolt must support onefourth of 25.0 kips or 6.25 kips/bolt.

ASD From the W1240 beam, each bolt must support onefourth of 16.7 kips or 4.18 kips/bolt.

From the W2150 beam, each bolt must support onetenth of 110 kips or 11.0 kips/bolt.

From the W2150 beam, each bolt must support onetenth of 73.3 kips or 7.33 kips/bolt.

The required strength for each of the shared bolts is: LRFD

ASD

Ru  6.25 kips/bolt  11.0 kips/bolt  17.3 kips/bolt

Ra  4.18 kips/bolt  7.33 kips/bolt  11.5 kips/bolt

From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear is: LRFD

ASD

rn  35.8 kips/bolt  17.3 kips/bolt o.k.

rn  23.9 kips/bolt  11.5 kips/bolt o.k. 

The available bearing and tearout strength of the girder web is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

ASD

rn

rn   87.8 kip/in. 0.520 in.  45.7 kips/bolt  17.3 kips/bolt o.k.



  58.5 kip/in. 0.520 in.  30.4 kips/bolt  11.5 kips/bolt

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Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-9 WEB)

OFFSET ALL-BOLTED DOUBLE-ANGLE CONNECTIONS (BEAMS-TO-GIRDER

Given:

Verify the all-bolted double-angle connections for back-to-back ASTM A992 W1645 beams to an ASTM A992 W3099 girder-web to support the end reactions shown in Figure II.A-9-1. The beam centerlines are offset 6 in. and the beam connections share a vertical row of bolts. Use ASTM A36 angles. The strength of the W1645 beams and angles are verified in Example II.A-4 and are not repeated here.

Fig. II.A-9-1. Connection geometry for Example II.A-9. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beams and girder ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Girder W1850 tw = 0.355 in. d = 18.0 in. Beam W1645

tw = 0.345 in. d = 16.1 in. Modify the 2L5324 SLBB connection designed in Example II.A-4 to work in the configuration shown in Figure II.A-9-1. The offset dimension (6 in.) is approximately equal to the gage on the support from the previous example (64 in.) and, therefore, is not recalculated. Thus, the available strength of the middle vertical row of bolts (through both connections) that carry a portion of the reaction for both connections must be verified for this new configuration. From ASCE/SEI 7, Chapter 2, the required strength of the Beam A and Beam B connections to the girder web is: LRFD Ru  1.2 10 kips   1.6  30 kips 

ASD

Ra  10 kips  30 kips  40.0 kips

 60.0 kips

In the girder web connection, each bolt will have the same effective strength; therefore, check the individual bolt effective strength. At the middle vertical row of bolts, the required strength for one bolt is the sum of the required shear strengths per bolt for each connection. LRFD  60.0 kips  ru   2 sides     6 bolts   20.0 kips/bolt (for middle vertical row)

ASD  40.0 kips  ra   2 sides     6 bolts   13.3 kips/bolt (for middle vertical row)

Bolt Shear From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear is: LRFD

rn  35.8 kips/bolt  20.0 kips/bolt o.k.

ASD rn  23.9 kips/bolt  13.3 kips/bolt o.k. 

Bearing on the Girder Web The available bearing strength per bolt is determined from AISC Manual Table 7-4 with s = 3 in. LRFD rn   87.8 kip/in. 0.355 in.  31.2 kips/bolt  20.0 kips/bolt o.k.

ASD rn   58.5 kip/in. 0.355 in.   20.8 kips/bolt  13.3 kips/bolt o.k.

Note: If the bolts are not spaced equally from the supported beam web, the force in each column of bolts should be determined by using a simple beam analogy between the bolts, and applying the laws of statics.

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Conclusion The connections are found to be adequate as given for the applied loads.

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EXAMPLE II.A-10 SKEWED DOUBLE BENT-PLATE CONNECTION (BEAM-TO-GIRDER WEB) Given:

Design the skewed double bent-plate connection between an ASTM A992 W1677 beam and ASTM A992 W2794 girder-web to support the following beam end reactions: RD = 13.3 kips RL = 40 kips Use 70-ksi electrodes and ASTM A36 plates. The final design is shown in Figure II.A-10-1.

Fig. II.A-10-1. Skewed double bent-plate connection (beam-to-girder web).

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Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam and girder ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1677 tw = 0.455 in. d = 16.5 in. Girder W2794

tw = 0.490 in. From AISC Specification Table J3.3, for d-in.-diameter bolts with standard holes: dh = , in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2 13.3 kips   1.6  40 kips 

 80.0 kips

ASD

Ra  13.3 kips  40 kips  53.3 kips

From Figure II.A-10-1(c), assign load to each vertical row of bolts by assuming a simple beam analogy between bolts and applying the principles of statics. LRFD Required strength for bent plate A: Ru =

80.0 kips  24 in.

6.00 in.  30.0 kips

ASD Required strength for bent plate A: Ra =

 53.3 kips  24 in.

6.00 in.  20.0 kips

Required strength for bent plate B:

Required strength for bent plate B:

Ru  80.0 kips  30.0 kips  50.0 kips

Ra  53.3 kips  20.0 kips  33.3 kips

Assume that the welds across the top and bottom of the plates will be 22 in. long, and that the load acts at the intersection of the beam centerline and the support face.

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While the welds do not coincide on opposite faces of the beam web and the weld groups are offset, the locations of the weld groups will be averaged and considered identical. See Figure II.A-10-1(d). Weld Design Assume a plate length of l = 82 in.

kl l 22 in.  82 in.  0.294

k

Interpolating from AISC Manual Table 8-8, with angle = 0, and k = 0.294, x = 0.0544

xl   0.0544  82 in.  0.462 in. a

 al  xl   xl

l 3s in  0.462 in.  82 in.  0.372

Interpolating from AISC Manual Table 8-8, with  = 0, a = 0.372, and k = 0.294, C = 2.52 The required weld size is determined as follows:   0.75

Dreq  

LRFD

Ru CC1l 50.0 kips 0.75  2.52 1.0  82 in.

 3.11 sixteenths

  2.00

Dreq  

ASD

Ra CC1l

2.00  33.3 kips 

2.52 1.0  82 in.

 3.11 sixteenths

Use 4-in. fillet welds and at least c-in.-thick bent plates to allow for the welds. Beam Web Strength at Fillet Weld The minimum beam web thickness required to match the shear rupture strength of the weld to that of the base metal is:

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t min  

6.19 Dmin Fu

(from Manual Eq. 9-3)

6.19  3.11

65 ksi  0.296 in.  0.455 in.

o.k .

Bolt Strength The effective strength of the individual fasteners is the lesser of the bolt shear strength per AISC Specification Section J3.6, and the bolt bearing and tearout strength per AISC Specification Section J3.10. By observation, the bent plate will govern over the girder web as it is thinner and lower strength material. Trying a c-in. plate the available strength at the critical vertical row of bolts (bent plate B) is determined as follows. From AISC Manual Table 7-1, the available shear strength per bolt for d-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in single shear is: LRFD

ASD rn  16.2 kips/bolt 

rn  24.3 kips/bolt

The available bearing and tearout strength of the bent-plate at the top edge bolt is determined using AISC Manual Table 7-5 with lev = 14 in. LRFD

rn   40.8 kip/in. c in.  12.8 kips/bolt

ASD rn   27.2 kip/in. c in.   8.50 kips/bolt

The available bearing and tearout strength of the bent-plate at the other bolts (not adjacent to the edge) is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

rn   91.4 kip/in. c in.  28.6 kips/bolt

ASD rn   60.9 kip/in. c in.   19.0 kips/bolt

The bolt shear strength governs over bearing and tearout for the other bolts (not adjacent to the edge); therefore, the effective strength of the bolt group is determined as follows: LRFD Rn  1 bolt 12.8 kips/bolt    2 bolts  24.3 kips/bolt   61.4 kips  50.0 kips o.k.

ASD Rn  1 bolt  8.50 kips/bolt     2 bolts 16.2 kips/bolt   40.9 kips  33.3 kips

o.k.

Shear Strength of Plate From AISC Specification Section J4.2, the available shear yielding strength of bent plate B (see Figure II.A-10-1) is determined as follows:

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Agv  lt   82 in. c in.  2.66 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  2.66 in.2



 57.5 kips

LRFD

  1.00

  1.50

Rn  1.00  57.5 kips 

ASD

Rn 57.5 kips   1.50  38.3 kips  33.3 kips o.k.

 57.5 kips  50.0 kips o.k.

From AISC Specification Section J4.2, the available shear rupture strength of bent plate B is determined as follows: Anv   l  n  d h  z in.  t  82 in.  3 , in. + z in.   c in.  1.72 in.2

Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  58 ksi  1.72 in.2



 59.9 kips

LRFD

  0.75

  2.00

Rn  0.75  59.9 kips 

ASD

Rn 59.9 kips   2.00  30.0 kips  33.3 kips n.g.

 44.9 kips  50.0 kips n.g.

Therefore, the plate thickness is increased to a in. The available shear rupture strength is: Anv   d  n , in. + z in.  t  8 2 in.  3 , in. + z in.   a in.  2.06 in.2

Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  58 ksi  2.06 in.

2



 71.7 kips

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LRFD

  0.75

ASD

  2.00

Rn  0.75  71.7 kips 

Rn 71.7 kips   2.00  35.9 kips  33.3 kips o.k.

 53.8 kips  50.0 kips o.k. Block Shear Rupture of Plate

The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the plate is determined as follows, using AISC Manual Tables 9-3a, 9-3b and 9-3c and AISC Specification Equation J4-5, with n = 3, lev = leh = 14 in., and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a: 

Fu Ant  32.6 kip/in.  t

ASD Tension rupture component from AISC Manual Table 9-3a:





Fu Ant  21.8 kip/in. t

Shear yielding component from AISC Manual Table 9-3b: 

Shear yielding component from AISC Manual Table 9-3b:





0.6Fy Agv  117 kip/in.  t

0.6Fy Agv  78.3 kip/in.  t





Shear rupture component from AISC Manual Table 9-3c:

Shear rupture component from AISC Manual Table 9-3c:



0.6Fu Anv  124 kip/in.  t



Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant



 124 kip/in.  32.6 kip/in. a in.  117 kip/in.  32.6 kip/in. a in.  58.7 kips  56.1 kips

0.6Fu Anv  82.6 kip/in. t Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant +      82.6 kip/in.  21.8 kip/in. a in.   78.3 kip/in.  21.8 kip/in. a in.  39.2 kips  37.5 kips

Therefore:

Therefore: Rn  56.1 kips  50.0 kips

o.k.

Rn  37.5 kips  33.3 kips o.k. 

Thus, the configuration shown in Figure II.A-10-1 can be supported using a-in. bent plates, and 4-in. fillet welds.

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EXAMPLE II.A-11A

SHEAR END-PLATE CONNECTION (BEAM-TO-GIRDER WEB)

Given:

Verify a shear end-plate connection to connect an ASTM A992 W1850 beam to an ASTM A992 W2162 girder web, as shown in Figure II.A-11A-1, to support the following beam end reactions: RD = 10 kips RL = 30 kips Use 70-ksi electrodes and ASTM A36 plate.

Fig. II.A-11A-1. Connection geometry for Example II.A-11A. Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam and girder ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1850 tw = 0.355 in. Girder W2162 tw = 0.400 in.

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From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2 10 kips   1.6  30 kips 

ASD

Ra  10 kips  30 kips  40.0 kips

 60.0 kips Bolt and End-Plate Available Strength

Tabulated values in AISC Manual Table 10-4 consider the limit states of bolt shear, bolt bearing on the end plate, tearout on the end plate, shear yielding of the end plate, shear rupture of the end plate, and block shear rupture of the end plate. From AISC Manual Table 10-4, for three rows of w-in.-diameter bolts and 4-in. plate thickness: LRFD

ASD Rn  50.9 kips  40.0 kips o.k. 

Rn  76.4 kips  60.0 kips o.k. 

Weld and Beam Web Available Strength Try x-in. weld. From AISC Manual Table 10-4, the minimum beam web thickness is: tw min  0.286 in.  0.355 in. o.k.

From AISC Manual Table 10-4, the weld and beam web available strength is: LRFD Rn  67.9 kips  60.0 kips o.k. 

ASD Rn  45.2 kips  40.0 kips o.k. 

Bolt Bearing on Girder Web From AISC Manual Table 10-4: LRFD Rn   527 kip/in. 0.400 in.

 211 kips  60.0 kips o.k.

ASD Rn   351 kip/in. 0.400 in.   140 kips  40.0 kips o.k.

Coped Beam Strength As was shown in Example II.A-4, the coped section does not control the design. o.k. Beam Web Shear Yielding As was shown in Example II.A-4, beam web shear does not control the design. o.k.

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EXAMPLE II.A-11B

END-PLATE CONNECTION SUBJECT TO AXIAL AND SHEAR LOADING

Given:

Verify the available strength of an end-plate connection for an ASTM A992 W18x50 beam, as shown in Figure II.A-11B-1, to support the following beam end reactions: LRFD Shear, Vu = 75 kips Axial tension, Nu = 60 kips

ASD Shear, Va = 50 kips Axial tension, Na = 40 kips

Use 70-ksi electrodes and ASTM A36 plate.

Fig. II.A-11B-1. Connection geometry for Example II.A-11B. Solution:

From AISC Manual Table 2-4 and 2-5, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1850 d = 18.0 in.

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tw = 0.355 in. Ag = 14.7 in.2 From AISC Specification Table J3.3, for d-in.-diameter bolts with standard holes: dh = , in. The resultant load is: LRFD 2

Ru  Vu  Nu 

ASD

2

2

Ra  Va  N a

 75 kips 2   60 kips 2



 96.0 kips

2

 50 kips 2   40 kips 2

 64.0 kips

The connection will first be checked for the shear load. The following bolt shear, bearing and tearout calculations are for a pair of bolts. Bolt Shear From AISC Manual Table 7-1, the available shear strength for d-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear, or pair of bolts in this example, is: LRFD

ASD rn  32.5 kips/pair of bolts 

rn  48.7 kips/pair of bolts Bolt Bearing on the Plate

The nominal bearing strength of the plate is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration: rn   2 bolts/row  2.4dtFu

(from Spec. Eq. J3-6a)

  2 bolts/row  2.4  d in.2 in. 58 ksi   122 kips (for a pair of bolts)

From AISC Specification Section J3.10, the available bearing strength of the plate for a pair of bolts is:   0.75

LRFD

rn  0.75 122 kips   91.5 kips/pair of bolts

  2.00

ASD

rn 122 kips  2.00   61.0 kips/pair of bolts

Bolt Tearout on the Plate The available tearout strength of the plate is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration. For the top edge bolts:

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lc  le  0.5d h  1 4 in.  0.5 , in.  0.781 in.

rn   2 bolts/row 1.2lc tFu

(from Spec. Eq. J3-6c)

  2 bolts/row 1.2  0.781 in.2 in. 58 ksi   54.4 kips (for a pair of bolts) The available bolt tearout strength for the pair of top edge bolts is:   0.75

LRFD

  2.00

rn  0.75  54.4 kips 

ASD

rn 54.4 kips   2.00  27.2 kips/pair of bolts

 40.8 kips/pair of bolts

Tearout controls over bolt shear and bearing strength for the top edge bolts in the plate. For interior bolts: lc  s  d h  3.00 in.  , in.  2.06 in.

rn   2 bolts/row 1.2lc tFu

(from Spec. Eq. J3-6c)

=  2 bolts/row 1.2  2.06 in.2 in. 58 ksi   143 kips/pair of bolts

The available bolt tearout strength for a pair of interior bolts is:   0.75

LRFD

  2.00

rn  0.75 143 kips 

ASD

rn 143 kips   2.00  71.5 kips/pair of bolts

 107 kips/pair of bolts

Bolt shear controls over tearout and bearing strength for the interior bolts in the plate. Shear Strength of Bolted Connection LRFD Rn  1 row  40.8 kips/pair of bolts 

  4 rows  48.7 kips/pair of bolts   236 kips  75 kips o.k.

ASD Rn = 1 row  27.2 kips/pair of bolts     4 rows  32.5 kips/pair of bolts   157 kips  50 kips

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Bolt Shear and Tension Interaction The available strength of the bolts due to the effect of combined tension and shear is determined from AISC Specification Section J3.7. The required shear stress is:

frv 

Vr nAb

where

Ab  0.601 in.2 (from AISC Manual Table 7-1) n  10 bolts LRFD f rv 



75 kips

10 0.601 in.2

ASD f rv 



 12.5 ksi



50 kips

10 0.601 in.2



 8.32 ksi

The nominal tensile stress modified to include the effects of shear stress is determined from AISC Specification Section J3.7 as follows. From AISC Specification Table J3.2:

Fnt  90 ksi Fnv  54 ksi LRFD

  0.75

Fnt  1.3Fnt 

Fnt f rv  Fnt Fnv

 1.3  90 ksi  

(Spec. Eq. J3-3a)

90 ksi 12.5 ksi   90 ksi 0.75  54 ksi 

 89.2 ksi  90 ksi

o.k .

ASD

  2.00

Fnt  1.3Fnt 

Fnt f rv  Fnt Fnv

 1.3  90 ksi  

2.00  90 ksi 

54 ksi  89.3 ksi  90 ksi o.k .

(Spec. Eq. J3-3b)

 8.32 ksi   90 ksi

Using the value of Fnt  89.2 ksi determined for LRFD, the nominal tensile strength of one bolt is: rn  Fnt Ab



  89.2 ksi  0.601 in.2

(Spec. Eq. J3-2)



 53.6 kips

The available tensile strength due to combined tension and shear is:   0.75

rn  0.75  53.6 kips   40.2 kips/bolt

LRFD

  2.00

rn 53.6 kips   2.00  26.8 kips/bolt

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LRFD

ASD Rn rn n    10 bolts  26.8 kips/bolt 

Rn  nrn  10 bolts  40.2 kips/bolt   402 kips  60 kips

o.k.

 268 kips  40 kips o.k.

Prying Action From AISC Manual Part 9, the available tensile strength of the bolts in the end-plate taking prying action into account is determined as follows: width of plate  gage 2 82 in.  52 in.  2  1.50 in.

a

Note: If a at the supporting element is smaller than a = 1.50 in., use the smaller a in the preceding calculations. gage  tw 2 52 in.  0.355 in.  2  2.57 in.

b

d  a   a  b 2 

db      1.25b   2    d in. d in.  1.50 in.   1.25  2.57 in.  2 2  1.94 in.  3.65 in.

(Manual Eq. 9-23)

 1.94 in.

d   b   b  b  2    2.57 in. 

(Manual Eq. 9-18) d in. 2

 2.13 in.

 

b a

(Manual Eq. 9-22)

2.13 in. 1.94 in.

 1.10

Note that end distances of 14 in. are used on the end-plate, so p is the average pitch of the bolts:

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l n

p 

142 in. 5

 2.90 in.

Check ps 2.90 in.  3 in.

o.k.

d   dh  , in. d p , in.  1 2.90 in.  0.677

  1

(Manual Eq. 9-20)

From AISC Manual Equations 9-26a or 9-26b, the required end-plate thickness to develop the available strength of the bolt without prying action is:   0.90 

LRFD

Bc  40.2 kips/bolt (calculated previously)

4 Bc b pFu

tc 

Bc  26.8 kips/bolt (calculated previously)

tc 

4  40.2 kips/bolt  2.13 in.



ASD

  1.67 



0.90  2.90 in. 58 ksi 

4 Bc b pFu 1.67  4  26.8 kips/bolt  2.13 in.

 2.90 in. 58 ksi 

1.51 in.

1.50 in.

Because the end-plate thickness of 2 in. is less than tc, using the value of tc = 1.51 in. determined for ASD, calculate the effect of prying action on the bolts.   tc  2  1    1  1     t    1.51 in. 2  1     1 0.677 1  1.10   2 in.    5.71

 

Because    1, the end-plate has insufficient strength to develop the bolt strength, therefore:

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(Manual Eq. 9-28)

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2

t  Q    1     tc  2

 2 in.    1  0.677   1.51 in.   0.184 The available tensile strength of the bolts taking prying action into account is determined from AISC Manual Equation 9-27 as follows: LRFD Tc  Bc Q

ASD Tc  Bc Q

  40.2 kips/bolt  0.186 

  26.8 kips/bolt  0.184 

 7.48 kips/bolt

 4.93 kips/bolt

Rn  Tc n   7.48 kips/bolt 10 bolts   74.8 kips  60 kips

o.k.

Rn  Tc n    4.93 kips/bolt 10 bolts   49.3 kips  40 kips

o.k.

Weld Design Assume a x-in. fillet weld on each side of the beam web, with the weld stopping short of the end of the plate at a distance equal to the weld size.

lw  142 in.  2  x in.  14.1 in. LRFD N    tan 1  u   Vu   60 kips  = tan 1    75 kips   38.7

ASD N    tan 1  a   Va   40 kips   tan 1    50 kips   38.7

From AISC Manual Table 8-4 for Angle = 30° (which will lead to a conservative result): Special Case: k  a  0 C  4.37

The required weld size is determined from AISC Manual Equation 8-21 as follows:

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LRFD

  0.75

Dmin  

  2.00

ASD

Ra CC1lw 2.00  64.0 kips   4.37  1.0 14.1 in.

Ru CC1lw

Dmin 

96.0 kips 0.75  4.37 1.0 14.1 in.

 2.08 sixteenths

 2.08 sixteenths

Use a x-in. fillet weld (minimum size from AISC Specification Table J2.4). Beam Web Strength at Fillet Weld The minimum beam web thickness required to match the shear rupture strength of the connecting element to that of the base metal is:

tmin  

6.19 Dmin Fu

(from Manual Eq. 9-3)

6.19  2.08 

65 ksi  0.198 in.  0.355 in.

o.k.

Shear Strength of the Plate From AISC Specification Section J4.2(a), the available shear yielding strength of the plate is determined as follows: Agv  2lt   2 142 in.2 in.  14.5 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  14.5 in.2



 313 kips

  1.00 

LRFD

  1.50 

Rn  1.00  313 kips   313 kips  96.0 kips

ASD

Rn 313 kips  1.50   209 kips  64.0 kips

o.k.

o.k .

From AISC Specification Section J4.2(b), the available shear rupture strength of the plate is determined as follows:

Anv  2 l  n  dh  z in.  t  2 142 in.  5 , in.  z in.  2 in.  9.50 in.2

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Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  58 ksi  9.50 in.

2



 331 kips

  0.75 

LRFD

  2.00 

Rn  0.75  331 kips   248 kips  96.0 kips

ASD

Rn 331 kips  2.00   166 kips  64.0 kips

o.k.

o.k.

Block Shear Rupture Strength of the Plate The nominal strength for the limit state of block shear rupture of the plate assuming an L-shaped tearout relative to shear load, is determined as follows. The tearout pattern is shown in Figure II.A-11B-2.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant where b  gage 2 82 in.  52 in.  2  1.50 in.

leh 

Agv   2  lev   n  1 s   t    2  14 in.   5  1 3.00 in.  2 in.  13.3 in.2

Fig. II.A-11B-2. Block shear rupture of end-plate.

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Anv  Agv   2  n  0.5 d h  z in. t   13.3 in.2 –  2  5  0.5, in.  z in.2 in.  8.80 in.2 Ant   2  leh  0.5  d h  z in.   t    2  1.50 in. – 0.5 , in.  z in.  2 in.

U bs

 1.00 in.2  1.0

and















Rn  0.60  58 ksi  8.80 in.2  1.0  58 ksi  1.00 in.2  0.60  36 ksi  13.3 in.2  1.0  58 ksi  1.00 in.2



 364 kips  345 kips

Therefore: Rn  345 kips

LRFD

  0.75 

Rn  0.75  345 kips   259 kips  75 kips

o.k.

  2.00 

Rn 345 kips   2.00  173 kips  50 kips

ASD

o.k .

Shear Strength of Beam From AISC Specification Section J4.2(a), the available shear yielding strength of the beam is determined as follows: Agv  dtw  18.0 in. 0.355 in.  6.39 in.2

Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  6.39 in.2



 192 kips

  1.00 

LRFD

Rn  1.00 192 kips   192 kips  75 kips

o.k.

  1.50 

Rn 192 kips   1.50  128 kips  50 kips

ASD

o.k .

The limit state of shear rupture of the beam web does not apply in this example because the beam is uncoped. Tensile Strength of Beam

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From AISC Specification Section J4.1, the available tensile yield strength of the beam is determined as follows: Rn  Fy Ag

(Spec. Eq. J4-1)



  50 ksi  14.7 in.

2



 735 kips

LRFD

  0.90 

  1.67 

Rn  0.90  735 kips   662 kips  60 kips

Rn 735 kips   1.67  440 kips  40 kips

o.k.

ASD

o.k.

From AISC Specification Section J4.1, determine the available tensile rupture strength of the beam. The effective net area is Ae  AnU from AISC Specification Section D3, where U is determined from AISC Specification Table D3.1, Case 3. U = 1.0

An  area of the directly connected elements  lw t w  14.1 in. 0.355 in.  5.01 in.2 The available tensile rupture strength is: Rn  Fu Ae  Fu AnU

(Spec. Eq. J4-2)





  65 ksi  5.01 in.2 1.0   326 kips

  0.75

LRFD

Rn  0.75  326 kips   245 kips  60 kips

o.k.

  2.00 

Rn 326 kips   2.00  163 kips  40 kips

Conclusion The connection is found to be adequate as given for the applied loads.

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ASD

o.k .

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EXAMPLE II.A-11C SHEAR END-PLATE CONNECTION—STRUCTURAL INTEGRITY CHECK Given:

Verify the shear end-plate connection from Example II.A-11B for the structural integrity provisions of AISC Specification Section B3.9. The ASTM A992 W1850 beam is bracing a column and the connection geometry is shown in Figure II.A-11C-1. Note that these checks are necessary when design for structural integrity is required by the applicable building code. Use 70-ksi electrodes and ASTM A36 plate.

Fig. II.A-11C-1. Connection geometry for Example II.A-11C. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W18x50

tw = 0.355 in.

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From Example II.A-11B, the required shear strength is: LRFD

ASD

Vu  75 kips  

Va  50 kips  

From AISC Specification Section B3.9, the required axial tensile strength is: LRFD 2 Tu  Vu  10 kips 3 2   75 kips   10 kips 3  50 kips  10 kips

ASD Ta  Va  10 kips  50 kips  10 kips  50 kips

 50 kips

From AISC Specification Section B3.9, these requirements are evaluated independently from other strength requirements. Bolt Tension From AISC Specification Section J3.6, the nominal bolt tensile strength is: Fnt = 90 ksi, from AISC Specification Table J3.2

Tn  nFnt Ab



 10 bolts  90 ksi  0.601 in.2

(from Spec. Eq. J3-1)



 541 kips Angle Bending and Prying Action From AISC Manual Part 9, the nominal strength of the end-plate accounting for prying action is determined as follows: width of plate  gage 2 82 in.  52 in.  2  1.50 in.

a

gage  t w 2 52 in.  0.355 in.  2  2.57 in.

b

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d   d   a    a  b   1.25b  b  2   2   d in. d in.  1.50 in.   1.25  2.57 in.  2 2  1.94 in.  3.65 in.

(Manual Eq. 9-23)

 1.94 in. b  b 

db 2

(Manual Eq. 9-18)

 2.57 in. 

d in. 2

 2.13 in. b a 2.13 in.  1.94 in.  1.10



(Manual Eq. 9-22)

Note that end distances of 14 in. are used on the end-plate, so p is the average pitch of the bolts: l n 142 in.  5  2.90 in.

p

Check p  s  3.00 in.

o.k.

d   dh , in. d p , in.  1 2.90 in.  0.677

  1

Bn  Fnt Ab



  90 ksi  0.601 in.2

(Manual Eq. 9-20)



 54.1 kips/bolt

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tc  

4 Bn b pFu

(from Manual Eq. 9-26)

4  54.1 kips/bolt  2.13 in.

 2.90 in. 58 ksi 

 1.66 in.   tc  2  1    1  1     t    1.66 in.  2  1     1 0.677 1  1.10   2 in.    7.05

 

(Manual Eq. 9-28)

Because   1, the end-plate has insufficient strength to develop the bolt strength, therefore: 2

t  Q    1     tc  2

 2 in.    1  0.677   1.66 in.   0.152 Tn  Bn Q

(from Manual Eq. 9-27)

 10 bolts  54.1 kips/bolt  0.152   82.2 kips

Weld Strength From AISC Specification Section J2.4, the nominal tensile strength of the weld is determined as follows:



Fnw  0.60 FEXX 1.0  0.50sin1.5 





(Spec. Eq. J2-5)

 0.60  70 ksi  1.0  0.50 sin1.5 90



 63.0 ksi

The weld length accounts for termination equal to the weld size. lw  l  2 w  142 in.  2  x in.  14.1 in.

The throat dimension is used to calculate the effective area of the fillet weld.

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w

lw  2 welds  2 x in.  14.1 in. 2 welds  2

Awe 

 3.74 in.2

Tn  Fnw Awe



  63.0 ksi  3.74 in.

2

(from Spec. Eq. J2-4)



 236 kips Tensile Strength of Beam Web at the Weld From AISC Specification Section J4.1, the nominal tensile strength of the beam web at the weld is: Ae  lw t w  14.1 in. 0.355 in.  5.01 in.2

Tn  Fu Ae



  65 ksi  5.01 in.

2

(Spec. Eq. J4-2)



 326 kips Nominal Tensile Strength The controlling nominal tensile strength, Tn, is the least of those previously calculated:

Tn  min 541 kips, 82.2 kips, 236 kips, 326 kips  82.2 kips

Tn  82.2 kips  50 kips

LRFD o.k.

Tn  82.2 kips  50 kips

ASD o.k.

Column Bracing From AISC Specification Section B3.9(c), the minimum axial tension strength for the connection of a member bracing a column is equal to 1% of two-thirds of the required column axial strength for LRFD and equal to 1% of the required column axial for ASD. These requirements are evaluated independently from other strength requirements. The maximum column axial force this connection is able to brace is determined as follows: LRFD

2  Tn  0.01 Pu    3 

ASD

Tn  0.01Pa  

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LRFD Solving for the column axial force:

ASD Solving for the column axial force:

3  Pu  100  Tn  2  3  100    82.2 kips  2  12, 300 kips

Pa  100Tn  100  82.2 kips   8, 220 kips

As long as the required column axial strength is less than or equal to Pu = 12,300 kips or Pa = 8,220 kips, this connection is an adequate column brace.

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EXAMPLE II.A-12A WEB)

ALL-BOLTED UNSTIFFENED SEATED CONNECTION (BEAM-TO-COLUMN

Given:

Verify the all-bolted unstiffened seated connection between an ASTM A992 W1650 beam and an ASTM A992 W1490 column web, as shown in Figure II.A-12A-1, to support the following end reactions: RD = 9 kips RL = 27.5 kips Use ASTM A36 angles.

Fig. II.A-12A-1. Connection geometry for Example II.A-12A-1. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi

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From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1650 tw = 0.380 in. d = 16.3 in. bf = 7.07 in. tf = 0.630 in. kdes = 1.03 in. Column W1490

tw = 0.440 in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  9 kips   1.6  27.5 kips 

ASD

Ra  9 kips  27.5 kips  36.5 kips

 54.8 kips Minimum Bearing Length

From AISC Manual Part 10, the minimum required bearing length, lb min, is the length of bearing required for the limit states of web local yielding and web local crippling on the beam, but not less than kdes. Using AISC Manual Equations 9-46a or 9-46b and AISC Manual Table 9-4, the minimum required bearing length for web local yielding is: LRFD

lb min 

ASD

Ru  R1  kdes R2

lb min 

54.8 kips  48.9 kips  1.03 in. 19.0 kip/in.  0.311 in.  1.03 in. 

Therefore, lb min  kdes =1.03 in.

Ra  R1 /   kdes R2 / 

36.5 kips  32.6 kips  1.03 in. 12.7 kip/in.  0.307 in.  1.03 in. 

Therefore, lb min  kdes =1.03 in.

For web local crippling, the maximum bearing length-to-depth ratio is determined as follows (including 4-in. tolerance to account for possible beam underrun): 3.25 in.  lb      d  max 16.3 in.  0.199  0.2

Using AISC Manual Equations 9-48a or 9-48b and AISC Manual Table 9-4, when

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LRFD Ru  R3 lb min  R4 54.8 kips  67.2 kips  5.79 kip/in.

ASD Ra  R3 /  lb min  R4 /  36.5 kips  44.8 kips  3.86 kip/in.

This results in a negative quantity; therefore,

This results in a negative quantity; therefore,

lb min  kdes  1.03 in.

lb min  kdes  1.03 in.

Connection Selection AISC Manual Table 10-5 includes checks for the limit states of shear yielding and flexural yielding of the outstanding angle leg. For an 8-in. angle length with a s-in. thickness, a 32-in. minimum outstanding leg, and conservatively using lb, req =1z in., from AISC Manual Table 10-5: LRFD

ASD Rn  59.9 kips  36.5 kips o.k. 

Rn  90.0 kips  54.8 kips o.k. 

The L64s (4-in. OSL), 8-in. long with 52-in. bolt gage, Connection Type B (four bolts), is acceptable. From the bottom portion of AISC Manual Table 10-5 for L6, with w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N), the available shear strength is: LRFD

ASD Rn  47.7 kips  36.5 kips o.k. 

Rn  71.6 kips  54.8 kips o.k. Bolt Bearing and Tearout on the Angle

Due to the presence and location of the bolts in the outstanding leg of the angle, tearout does not control. The nominal bearing strength of the angles is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration: Rn   4 bolts  2.4dtFu

(from Spec. Eq. J3-6a)

  4 bolts  2.4  w in. s in. 58 ksi   261 kips   0.75

LRFD

Rn  0.75  261 kips   196 kips  54.8 kips o.k.

  2.00

ASD

Rn 261 kips   2.00  131 kips  36.5 kips

Note that the effective strength of the bolt group is controlled by bolt shear.

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Bolt Bearing on the Column The nominal bearing strength of the column web determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration, is: Rn   4 bolts  2.4dtw Fu

(from Spec. Eq. J3-6a)

  4 bolts  2.4  w in. 0.440 in. 65 ksi   206 kips   0.75

LRFD

Rn  0.75  206 kips   155 kips  54.8 kips o.k.

  2.00

ASD

Rn 206 kips   2.00  103 kips  36.5 kips

o.k.

Top Angle and Bolts As discussed in AISC Manual Part 10, use an L444 with two w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) through each leg. Conclusion The connection design shown in Figure II.A-12A-1 is acceptable.

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EXAMPLE II.A-12B ALL-BOLTED UNSTIFFENED SEATED CONNECTION—STRUCTURAL INTEGRITY CHECK Given:

Verify the all-bolted unstiffened seated connection from Example II.A-12A, as shown in Figure II.A-12B-1, for the structural integrity provisions of AISC Specification Section B3.9. The connection is verified as a beam and girder end connection and as an end connection of a member bracing a column. Note that these checks are necessary when design for structural integrity is required by the applicable building code. The beam is an ASTM A992 W1650 and the angles are ASTM A36 material.

Fig. II.A-12B-1. Connection geometry for Example II.A-12B. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Angle ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Beam W16x50

bf = 7.07 in. tf = 0.630 in. From AISC Specification Table J3.3, the hole diameter for w-in.-diameter bolts with standard holes is: dh = m in. From Example II.A-12A, the required shear strength is: LRFD

ASD

Vu  54.8 kips

Va  36.5 kips

From AISC Specification Section B3.9(b), the required axial tensile strength is: LRFD 2 Tu  Vu  10 kips 3 2   54.8 kips   10 kips 3  36.5 kips

ASD Ta  Va  10 kips  36.5 kips  10 kips  36.5 kips

From AISC Specification Section B3.9, these strength requirements are evaluated independently from other strength requirements. Bolt Shear Bolt shear is checked for the outstanding leg of the seat angle. From AISC Specification Section J3.6, the nominal bolt shear strength is: Fnv = 54 ksi, from AISC Specification Table J3.2

Tn  nFnv Ab



  2 bolts  54 ksi  0.442 in.

2

(from Spec. Eq. J3-1)



 47.7 kips Bolt Tension Bolt tension is checked for the top row of bolts on the support leg of the seat angle. From AISC Specification Section J3.6, the nominal bolt tensile strength is: Fnt = 90 ksi, from AISC Specification Table J3.2 Tn  nFnt Ab

(from Spec. Eq. J3-1)



  2 bolts  90 ksi  0.442 in.2



 79.6 kips

Bolt Bearing and Tearout

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Bolt bearing and tearout is checked for the outstanding leg of the seat angle. From AISC Specification Section B3.9, for the purpose of satisfying structural integrity requirements, inelastic deformations of the connection are permitted; therefore, AISC Specification Equations J3-6b and J3-6d are used to determine the nominal bearing and tearout strength. By inspection, bolt bearing and tearout will control for the angle. For bolt bearing on the angle: Tn  n3.0dtFu

(from Spec. Eq. J3-6b)

  2 bolts  3.0  w in. s in. 58 ksi   163 kips

For bolt tearout on the angle:

lc  leg  22 in.  0.5d h  4.00 in.  22 in.  0.5 m in.  1.09 in. Tn  n1.5lc tFu

(from Spec. Eq. J3-6d)

  2 bolts 1.5 1.09 in. s in. 58 ksi   119 kips

Angle Bending and Prying Action From AISC Manual Part 9, the nominal strength of the angle accounting for prying action is determined as follows: b  24 in. 

s in. 2

 1.94 in. a  min 2.50 in., 1.25b  min 2.50 in., 1.25 1.94 in.  2.43 in.

d   b   b  b  2    1.94 in. 

(Manual Eq. 9-18) w in. 2

 1.57 in.

a  a 

db 2

 2.43 in. 

(from Manual Eq. 9-23) w in. 2

 2.81 in.

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b a 1.57 in.  2.81 in.  0.559



(Manual Eq. 9-22)

Note that end distances of 14 in. are used on the angles, so p is the average pitch of the bolts: l n 8.00 in.  2  4.00 in.

p

Check p  s  52 in. o.k. d   dh m in.

d p m in.  1 4.00 in.  0.797

  1

Bn  Fnt Ab



  90 ksi  0.442 in.2

(Manual Eq. 9-20)



 39.8 kips/bolt

tc  

4 Bn b pFu

(from Manual Eq. 9-26)

4  39.8 kips/bolt 1.57 in.

 4.00 in. 58 ksi 

 1.04 in.   

  tc  2  1    1  1     t  

(Manual Eq. 9-28)

 1.04 in.  2  1    1 0.797 1  0.559   s in.  

 1.42

Because    1, the angle has insufficient strength to develop the bolt strength, therefore:

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2

t  Q    1     tc  2

 s in.    1  0.797   1.04 in.   0.649 Tn  Bn Q

(from Manual Eq. 9-27)

  2 bolts  39.8 kips/bolt  0.649   51.7 kips

Block Shear Rupture By comparison of the seat angle length and flange width, block shear rupture of the beam flange will control. The block shear rupture failure path is shown in Figure II.A-12B-2. From AISC Specification Section J4.3, the available block shear rupture strength of the beam flange is determined as follows (account for a possible 4-in. beam underrun): Tn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

where Agv   2  le t f   2 1w in. 0.630 in.  2.21 in.2 Anv   2   le  0.5  d h  z in.  t f   2  1w in.  0.5 m in.  z in.   0.630 in.  1.65 in.2

 b f  gage  Ant   2    0.5  d h  z in.  t f 2    7.07 in.  52 in.    2   0.5 m in.  z in.   0.630 in. 2    0.438 in.2

Fig. II.A-12B-2. Beam flange block shear rupture.

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U bs  1.0

and















Tn  0.60  65 ksi  1.65 in.2  1.0  65 ksi  0.438 in.2  0.60  50 ksi  2.21 in.2  1.0  65 ksi  0.438 in.2



 92.8 kips  94.8 kips  92.8 kips

Nominal Tensile Strength The controlling tensile strength, Tn, is the least of those previously calculated:

Tn  min 47.7 kips, 79.6 kips, 163 kips, 119 kips, 51.7 kips, 92.8 kips  47.7 kips LRFD Tn  47.7 kips  36.5 kips o.k.

ASD Tn  47.7 kips  36.5 kips o.k.

Column Bracing From AISC Specification Section B3.9(c), the minimum axial tension strength for the connection of a member bracing a column is equal to 1% of two-thirds of the required column axial strength for LRFD and equal to 1% of the required column axial for ASD. These requirements are evaluated independently from other strength requirements. The maximum column axial force this connection is able to brace is determined as follows, LRFD

ASD

2  Tn  0.01 Pu  3 

Tn  0.01Pa

Solving for the column axial force:

Solving for the column axial force:

3  Pu  100  Tn  2  3  100    47.7 kips  2  7,160 kips

Pa  100Tn  100  47.7 kips   4, 770 kips

As long as the required column axial strength is less than Pu = 7,160 kips or Pa = 4,770 kips, this connection is an adequate column brace.

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EXAMPLE II.A-13 BOLTED/WELDED COLUMN FLANGE)

UNSTIFFENED

SEATED

CONNECTION

(BEAM-TO-

Given:

Verify the unstiffened seated connection between an ASTM A992 W2162 beam and an ASTM A992 W1461 column flange, as shown in Figure II.A-13-1, to support the following beam end reactions: RD = 9 kips RL = 27.5 kips Use ASTM A36 angles and 70-ksi weld electrodes.

Fig. II.A-13-1. Connection geometry for Example II.A-13. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Beam W2162 tw = 0.400 in. d = 21.0 in. bf = 8.24 in. tf = 0.615 in. kdes= 1.12 in. Column W1461

tf = 0.645 in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  9 kips   1.6  27.5 kips 

ASD

Ra  9 kips  27.5 kips  36.5 kips

 54.8 kips Minimum Bearing Length

From AISC Manual Part 10, the minimum required bearing length, lb min, is the length of bearing required for the limit states of web local yielding and web local crippling on the beam, but not less than kdes. Using AISC Manual Equations 9-46a or 9-46b and AISC Manual Table 9-4, the minimum required bearing length for web local yielding is: LRFD

ASD

Ru  R1  kdes R2 54.8 kips  56.0 kips   1.12 in. 20.0 kip/in. which results in a negative quantity.

Ra  R1 /   kdes R2 /  36.5 kips  37.3 kips   1.12 in. 13.3 kip/in. which results in a negative quantity.

Therefore, lb min  k des  1.12 in.

Therefore, lb min  k des  1.12 in.

lb min 

lb min 

For web local crippling, the maximum bearing length-to-depth ratio is determined as follows (including a 4-in. tolerance to account for possible beam underrun): 3.25 in.  lb     d   max 21.0 in.  0.155  0.2

From AISC Manual Equations 9-48a or 9-48b and AISC Manual Table 9-4, when LRFD

Ru  R3 R4 54.8 kips  71.7 kips  5.37 kip/in.

lb min 

lb  0.2 : d ASD

Ra  R3 /  R4 /  36.5 kips  47.8 kips  3.58 kip/in.

lb min 

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LRFD This results in a negative quantity; therefore,

ASD This results in a negative quantity; therefore,

lb min  k des  1.12 in.

lb min  k des  1.12 in.

Connection Selection AISC Manual Table 10-6 includes checks for the limit states of shear yielding and flexural yielding of the outstanding angle leg. For an 8-in. angle length with a s-in. thickness, a 32-in. minimum outstanding leg, and conservatively using lb, req = 18 in., from AISC Manual Table 10-6: LRFD

Rn  81.0 kips  54.8 kips o.k. 

ASD Rn  53.9 kips  36.5 kips o.k. 

From AISC Manual Table 10-6, for a L84s (4-in. OSL), 8-in. long, with c-in. fillet welds, the weld available strength is: LRFD

Rn  66.7 kips  54.8 kips o.k.

ASD Rn  44.5 kips  36.5 kips o.k. 

Use two w-in.-diameter bolts with threads not excluded from the shear plane (thread condition N) to connect the beam to the seat angle. The strength of the bolts, welds and angles must be verified if horizontal forces are added to the connection. Top Angle, Bolts and Welds Use an L444 with two w-in.-diameter bolts with threads not excluded from the shear plane (thread condition N) through the supported beam leg of the angle. Use a x-in. fillet weld along the toe of the angle to the column flange. See the discussion in AISC Manual Part 10. Conclusion The connection design shown in Figure II.A-13-1 is acceptable.

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EXAMPLE II.A-14 BOLTED/WELDED STIFFENED SEATED CONNECTION (BEAM-TO-COLUMN FLANGE) Given:

Verify the bolted/welded stiffened seated connection between an ASTM A992 W2168 beam and an ASTM A992 W1490 column flange, as shown in Figure II.A-14-1, to support the following end reactions: RD = 21 kips RL = 62.5 kips Use 70-ksi weld electrodes and ASTM A36 angles and plate.

Fig. II.A-14-1. Connection geometry for Example II.A-14. Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Angle and plates ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Beam W2168 tw = 0.430 in. d = 21.1 in. bf = 8.27 in. tf = 0.685 in. kdes = 1.19 in. Column W1490

tf = 0.710 in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  21 kips   1.6  62.5 kips 

ASD

Ra  21 kips  62.5 kips  83.5 kips

 125 kips Required Stiffener Width

The minimum stiffener width, Wmin, is determined based on limit states of web local yielding and web local crippling for the beam. The minimum stiffener width for web local crippling of the beam web, for the force applied less than one-half of the depth of the beam from the end of the beam and assuming lb/d > 0.2, is determined from AISC Manual Equations 949a or 9-49b and AISC Manual Table 9-4, as follows (including a 4-in. tolerance to account for possible beam underrun):

Wmin

LRFD Ru  R5   setback  underrun R6 125 kips  75.9 kips   2 in.  4 in. 7.95 kip/in.  6.93 in.

ASD

Ra  R5 /   setback  underrun R6 /  83.5 kips  50.6 kips   2 in.  4 in. 5.30 kip/in.  6.96 in.

Wmin 

The minimum stiffener width for web local yielding of the beam, for the force applied less than the depth of the beam from the end of the beam, is determined from AISC Manual Equations 9-46a or 9-46b and AISC Manual Table 9-4, as follows (including a 4-in. tolerance to account for possible beam underrun):

Wmin

LRFD Ru  R1   setback  underrun R2 125 kips  64.0 kips   2 in.  4 in. 21.5 kip/in.  3.59 in.

Wmin

ASD Ra  R1 /    setback  underrun R2 /  83.5 kips  42.6 kips   2 in.  4 in. 14.3 kip/in.  3.61 in.

Use W = 7 in. Check assumption:

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lb 6.25 in.  d 21.1 in.  0.296  0.2 o.k. Stiffener Length and Stiffene- to-Column Flange Weld Size Use a stiffener with l = 15 in. and c-in. fillet welds. From AISC Manual Table 10-8, with W = 7 in.: LRFD

Rn  139 kips > 125 kips o.k. 

ASD Rn  93.0 kips > 83.5 kips o.k. 

Seat Plate Welds Use c-in. fillet welds on each side of the stiffener. From AISC Manual Figure 10-10(b), minimum length of seat plate-to-column flange weld is 0.2(L) = 3 in. As discussed in AISC Manual Part 10, the weld between the seat plate and stiffener plate is required to have a strength equal to or greater than the weld between the seat plate and the column flange, use c-in. fillet welds on each side of the stiffener to the seat plate; length of weld = 6 in. per side. Seat Plate Dimensions A dimension of 9 in. is adequate to accommodate the w-in.-diameter bolts on a 52-in. gage connecting the beam flange to the seat plate. Use a PLa in.7 in.9 in. for the seat. Stiffener Plate Thickness As discussed in AISC Manual Part 10, the minimum stiffener plate thickness to develop the seat plate weld for Fy = 36 ksi plate material is:

tmin  2w  2  c in.  s in. As discussed in AISC Manual Part 10, the minimum plate thickness for a stiffener with Fy = 36 ksi and a beam with Fy = 50 ksi is:

 50 ksi  tmin    tw  36 ksi   50 ksi     0.430 in.  36 ksi   0.597 in.  s in. Use a PLs in.7 in.1 ft 3 in. Top Angle, Bolts and Welds

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Use an L444with two w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) through the supported beam leg of the angle. Use a x-in. fillet weld along the toe of the angle to the column flange. See discussion in AISC Manual Part 10. Conclusion The connection design shown in Figure II.A-14-1 is acceptable.

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EXAMPLE II.A-15 BOLTED/WELDED STIFFENED SEATED CONNECTION (BEAM-TO-COLUMN WEB) Given:

Verify the stiffened seated connection between an ASTM A992 W2168 beam and an ASTM A992 W1490 column web, as shown in Figure II.A-15-1, to support the following beam end reactions: RD = 21 kips RL = 62.5 kips Use 70-ksi weld electrodes and ASTM A36 angles and plate.

Fig. II.A-15-1. Connection geometry for Example II.A-15. Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Angle and Plates ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Beam W2168 tw = 0.430 in. d = 21.1 in. bf = 8.27 in. tf = 0.685 in. kdes = 1.19 in. Column W1490

tw = 0.440 in. T = 10 in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  21 kips   1.6  62.5 kips 

ASD

Ra  21 kips  62.5 kips  83.5 kips

 125 kips Required Stiffener Width

The minimum stiffener width, Wmin, is determined based on limit states of web local yielding and web local crippling for the beam. The minimum stiffener width for web local crippling of the beam web, for the force applied less than one-half of the depth of the beam from the end of the beam and assuming lb/d > 0.2, is determined from AISC Manual Equations 9-49a or 9-49b and AISC Manual Table 9-4, as follows (including a 4-in. tolerance to account for possible beam underrun):

Wmin

LRFD Ru  R5   setback  underrun R6 125 kips  75.9 kips   2 in.  4 in. 7.95 kip/in.  6.93 in.

Wmin

ASD Ra  R5 /    setback  underrun R6 /  83.5 kips  50.6 kips   2 in.  4 in. 5.30 kip/in.  6.96 in.

The minimum stiffener width for web local yielding of the beam, for the force applied less than the depth of the beam from the end of the beam, is determined from AISC Manual Equations 9-46a or 9-46b and AISC Manual Table 9-4, as follows (including a 4-in. tolerance to account for possible beam underrun):

Wmin

LRFD Ru  R1   setback  underrun R2 125 kips  64.0 kips   2 in.  4 in. 21.5 kip/in.  3.59 in.

Wmin

ASD Ra  R1 /    setback  underrun R2 /  83.5 kips  42.6 kips   2 in.  4 in. 14.3 kip/in.  3.61 in.

Use W = 7 in. Check assumption:

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lb 6.25 in.  d 21.1 in.  0.296  0.2 o.k. Stiffener Length and Stiffener to Column Flange Weld Size Use a stiffener with l = 15 in. and c-in. fillet welds. From AISC Manual Table 10-8, with W = 7 in., the weld available strength is: LRFD

Rn  139 kips > 125 kips o.k. 

ASD Rn  93.0 kips > 83.5 kips o.k. 

Seat Plate Welds Use c-in. fillet welds on each side of the stiffener. From AISC Manual Figure 10-10(b), minimum length of seat plate-to-column flange weld is 0.2(L) = 3 in. As discussed in AISC Manual Part 10, the weld between the seat plate and stiffener plate is required to have a strength equal to or greater than the weld between the seat plate and the column flange, use c-in. fillet welds on each side of the stiffener to the seat plate; length of weld = 6 in. per side. Seat Plate Dimensions A dimension of 9 in. is adequate to accommodate the w-in.-diameter bolts on a 52-in. gage connecting the beam flange to the seat plate. Use a PLa in.7 in.9 in. for the seat. Stiffener Plate Thickness As discussed in AISC Manual Part 10, the minimum stiffener plate thickness to develop the seat plate weld for Fy = 36 ksi plate material is:

tmin  2w  2  c in.  s in. As discussed in AISC Manual Part 10, the minimum plate thickness for a stiffener with Fy = 36 ksi and a beam with Fy = 50 ksi is:

 50 ksi  tmin    tw  36 ksi   50 ksi     0.430 in.  36 ksi   0.597 in.  s in. Use a PLs in.7 in.1 ft 3 in. Top Angle, Bolts and Welds

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Use an L444with two w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) through the supported beam leg of the angle. Use a x-in. fillet weld along the toe of the angle to the column web. See discussion in AISC Manual Part 10. Column Web If the seat is welded to a column web, the base metal strength of the column must be checked. If only one side of the column web has a stiffened seated connection, then:

tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  5 sixteenths 

65 ksi  0.238 in.  0.440 in. o.k. If both sides of the column web have a stiffened seated connection, then:

tmin  

6.19 D Fu

(Manual Eq. 9-3)

6.19  5 sixteenths 

65 ksi  0.476 in.  0.440 in. n.g. The column is sufficient for a one-sided stiffened seated connection. For a two-sided connection the weld available strength must be reduced as discussed in AISC Manual Part 10. Note: Additional detailing considerations for stiffened seated connections are given in Part 10 of the AISC Manual. Conclusion The connection design shown in Figure II.A-15-1 is acceptable.

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EXAMPLE II.A-16 OFFSET UNSTIFFENED SEATED CONNECTION (BEAM-TO-COLUMN FLANGE) Given:

Verify the seat angle and weld size required for the unstiffened seated connection between an ASTM A992 W1438 beam and an ASTM A992 W1265 column flange connection with an offset of 52 in., as shown in Figure II.A-161, to support the following beam end reactions: RD = 5 kips RL = 15 kips Use an ASTM A36 angle and 70-ksi weld electrodes.

Fig. II.A-16-1. Connection geometry for Example II.A-16. Solution:

From AISC Manual Tables 2-4, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Angle ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1438 d = 14.1 in. kdes= 0.915 in.

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Column W1265 tf = 0.605 in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  5 kips   1.6 15 kips 

ASD

Ra  5 kips  15 kips  20.0 kips

 30.0 kips Minimum Bearing Length

From AISC Manual Part 10, the minimum required bearing length, lb min, is the length of bearing required for the limit states of web local yielding and web local crippling on the beam, but not less than kdes. From AISC Manual Equations 9-46a or 9-46b and AISC Manual Table 9-4, the minimum required bearing length for web local yielding is:

lb min

LRFD Ru  R1   kdes R2 30.0 kips  35.5 kips   0.915 in. 15.5 kip/in.

lb min

ASD   R R1 /  a  kdes R2 /  20.0 kips  23.6 kips   0.915 in. 10.3 kip/in.

This results in a negative quantity; therefore,

This results in a negative quantity; therefore,

lb min  k des  0.915 in.

lb min  k des  0.915 in.

From AISC Manual Equations 9-48a or 9-48b and AISC Manual Table 9-4, the minimum required bearing length for web local crippling, assuming lb d  0.2, is: LRFD

ASD

Ru  R3  kdes R4 30.0 kips  44.7 kips   0.915 in. 4.45 kip/in.

lb min 

lb min 



Ra  R3 /   kdes R4 / 

20.0 kips  29.8 kips  0.915 in. 2.96 kip/in.

This results in a negative quantity; therefore,

This results in a negative quantity; therefore,

lb min  k des  0.915 in.

lb min  k des  0.915 in.

Check assumption:

lb 0.915 in.  d 14.1 in.  0.0649  0.2 o.k. Seat Angle and Welds The required strength for the righthand weld can be determined by summing moments about the lefthand weld.

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LRFD

RuR 

 30.0 kips  3.00 in.

3.50 in.  25.7 kips

ASD

RaR 

 20.0 kips  3.00 in.

3.50 in.  17.1 kips

Conservatively design the seat for twice the force in the more highly loaded weld. Therefore, design the seat for the following: Ru  2  25.7 kips 

LRFD

 51.4 kips

Ra  2 17.1 kips 

ASD

 34.2 kips

Use a 6-in. angle length with a s-in. thickness and a 32-in. minimum outstanding leg and conservatively using lb, req = , in., from AISC Manual Table 10-6: LRFD

Rn  81.0 kips > 51.4 kips o.k. 

ASD Rn  54.0 kips  34.2 kips o.k. 

Use an L74s (4-in. OSL), 6-in. long with c-in. fillet welds. From AISC Manual Table 10-6, the weld available strength is: LRFD

Rn  53.4 kips  51.4 kips o.k.

ASD Rn  35.6 kips  34.2 kips o.k. 

Use an L74s0 ft 6 in. for the seat angle. Use two w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) to connect the beam to the seat angle. Weld the angle to the column with c-in. fillet welds. Top Angle, Bolts and Welds Use an L444 with two w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) through the outstanding leg of the angle. Use a x-in. fillet weld along the toe of the angle to the column flange [maximum size permitted by AISC Specification Section J2.2b(b)(2)]. Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-17A FLANGE)

SINGLE-PLATE CONNECTION (CONVENTIONAL BEAM-TO-COLUMN

Given:

Verify a single-plate connection between an ASTM A992 W1650 beam and an ASTM A992 W1490 column flange, as shown in Figure II.A-17A-1, to support the following beam end reactions: RD = 8 kips RL = 25 kips Use 70-ksi electrodes and an ASTM A36 plate.

Fig. II.A-17A-1. Connection geometry for Example II.A-17A. Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Beam W1650 tw = 0.380 in. d = 16.3 in. Column W1490 tf = 0.710 in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  8 kips   1.6  25 kips 

ASD

Ra  8 kips  25 kips  33.0 kips

 49.6 kips Connection Selection

AISC Manual Table 10-10a includes checks for the limit states of bolt shear, bolt bearing on the plate, tearout on the plate, shear yielding of the plate, shear rupture of the plate, block shear rupture of the plate, and weld shear. Use four rows of w-in.-diameter bolts in standard holes, 4-in. plate thickness, and x-in. fillet weld size. From AISC Manual Table 10-10a, the bolt, weld and single-plate available strength is: LRFD

ASD Rn  34.8 kips  33.0 kips o.k. 

Rn  52.2 kips  49.6 kips o.k.  Bolt Bearing and Tearout for Beam Web

Similar to the discussion in AISC Manual Part 10 for conventional, single-plate shear connections, the bearing and tearout are checked in accordance with AISC Specification Section J3.10, assuming the reaction is applied concentrically. The available bearing and tearout strength of the beam web is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

ASD

Rn   4 bolts  87.8 kip/in. 0.380 in.  134 kips  49.6 kips o.k.

Rn   4 bolts  58.5 kip/in. 0.380 in.   88.9 kips  33.0 kips o.k.

Note: To provide for stability during erection, it is recommended that the minimum plate length be one-half the Tdimension of the beam to be supported. AISC Manual Table 10-1 may be used as a reference to determine the recommended maximum and minimum connection lengths for a supported beam. Block shear rupture, shear yielding, and shear rupture will not control for an uncoped section. Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-17B SINGLE-PLATE CONNECTION SUBJECT TO AXIAL AND SHEAR LOADING (BEAM-TO-COLUMN FLANGE) Given:

Verify the available strength of a single-plate connection for an ASTM A992 W1850 beam connected to an ASTM A992 W1490 column flange, as shown in Figure II.A-17B-1, to support the following beam end reactions: LRFD Shear, Vu = 75 kips Axial tension, Nu = 60 kips

ASD Shear, Va = 50 kips Axial tension, Na = 40 kips

Use 70-ksi electrodes and an ASTM A572 Grade 50 plate.

Fig. II.A-17B-1. Connection geometry for Example II.A-17B. Solution:

From AISC Manual Table 2-4 and Table 2-5, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A572 Grade 50 Fy = 50 ksi Fu = 65 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1850 Ag = 14.7 in.2 d = 18.0 in.

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tw = 0.355 in. tf = 0.570 in. Column W1490 tf = 0.710 in. From AISC Specification Table J3.3, for d-in.-diameter bolts with standard holes: dh = , in. The resultant load is: LRFD 2

Ru  Vu  Nu 

ASD

2

2

Ra  Va  N a

 75 kips 2   60 kips 2



 96.0 kips

2

 50 kips 2   40 kips 2

 64.0 kips

The resultant load angle, measured from the vertical, is: LRFD  60 kips    tan 1    75 kips   38.7

ASD  40 kips    tan 1    50 kips   38.7

Bolt Shear Strength From AISC Manual Table 10-9, for single-plate shear connections with standard holes and n = 5: a 2 22 in.  2  1.25 in.

e

The coefficient for eccentrically loaded bolts is determined by interpolating from AISC Manual Table 7-6 for Angle = 30, n = 5 and ex = 1.25 in. Note that 30 is used conservatively in order to employ AISC Manual Table 7-6. A direct analysis method can be performed to obtain a more precise value using the instantaneous center of rotation method. C  4.60

From AISC Manual Table 7-1, the available shear strength for a d-in.-diameter Group A bolt with threads not excluded from the shear plane (thread condition N) is: LRFD rn  24.3 kips/bolt

ASD rn  16.2 kips/bolt 

Bolt Bearing on the Beam Web

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Note that bolt bearing and tearout of the beam web will control over bearing and tearout of the plate because the beam web is thinner and has less edge distance than the plate; therefore, those limit states will only be checked on the beam web. The nominal bearing strength is determined using AISC Specification Equation J3-6b in lieu of Equation J3-6a, because plowing of the bolts in the beam web is desirable to provide some flexibility in the connection. (Spec. Eq. J3-6b)

rn  3.0 dtFu  3.0  d in. 0.355 in. 65 ksi   60.6 kips/bolt

From AISC Specification Section J3.10, the available bearing strength of the beam per bolt is:   0.75

LRFD

rn  0.75  60.6 kips/bolt   45.5 kips/bolt

  2.00

ASD

rn 60.6 kips/bolt   2.00  30.3 kips/bolt

Bolt Tearout on the Beam Web The nominal tearout strength is determined using AISC Specification Equation J3-6d in lieu of Equation J3-6c, because plowing of the bolts in the beam web is desirable to provide some flexibility in the connection. Because the direction of the load on the bolt is unknown, the minimum bolt edge distance is used to determine a worst case available tearout strength. The bolt edge distance for the web in the horizontal direction controls for this design. If a computer program is available, the true lc can be calculated based on the instantaneous center of rotation. Therefore, for worst case edge distance in the beam web, and considering possible length underrun of 4 in. on the beam length: lc  leh  0.5d h  underrun  1w in.  0.5 , in.  4 in.  1.03 in.

rn  1.5lc tFu

(Spec. Eq. J3-6d)

 1.5 1.03 in. 0.355 in. 65 ksi   35.7 kips/bolt

  0.75

LRFD

rn  0.75  35.7 kips   26.8 kips/bolt

  2.00

ASD

rn 35.7 kips   2.00  17.9 kips/bolt

Strength of Bolted Connection Bolt shear is the controlling limit state for all bolts at the connection to the beam web. The available strength of the connection is:

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LRFD

ASD Rn Crn     4.60(16.2 kips/bolt)  74.5 kips > 64.0 kips o.k.

Rn  C rn  4.60  24.3 kips/bolt   112 kips > 96.0 kips

o.k.

Strength of Weld From AISC Manual Part 10, a weld size of (s)tp is used to develop the strength of the shear plate, because, in general, the moment generated by this connection is indeterminate.

w  st p  s 2 in.  c in. Use a two-sided c-in. fillet weld. Shear Strength of Supporting Column Flange From AISC Specification Section J4.2(b), the available shear rupture strength of the column flange is determined as follows: Anv   2 shear planes  lt f   2 shear planes 14.5 in. 0.710 in.  20.6 in.2 Rn  0.60 Fu Anv



 0.60  65 ksi  20.6 in.

2

(Spec. Eq. J4-4)



 803 kips

  0.75

LRFD

Rn  0.75  803 kips   602 kips  75 kips

o.k.

  2.00

Rn 803 kips   2.00  402 kips  50 kips

ASD

o.k.

The available shear yielding strength of the column flange need not be checked because Anv = Agv and shear rupture will control. Shear Yielding Strength of the Plate From AISC Specification Section J4.2(a), the available shear yielding strength of the plate is determined as follows: Agv  lt  14.5 in.2 in.  7.25 in.2

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Rn  0.60 Fy Agv



 0.60  50 ksi  7.25 in.

2

(Spec. Eq. J4-3)



 218 kips

LRFD

  1.00

Rn  1.00  218 kips   218 kips  75 kips

o.k.

  1.50

ASD

Rn 218 kips  1.50   145 kips  50 kips

o.k.

Tensile Yielding Strength of the Plate From AISC Specification Section J4.1(a), the available tensile yielding strength of the plate is determined as follows:

Ag  lt  14.5 in.2 in.  7.25 in.2 Rn  Fy Ag



  50 ksi  7.25 in.

2

(Spec. Eq. J4-1)



 363 kips LRFD

  0.90

Rn  0.90  363 kips   327 kips  60 kips

o.k.

  1.67

ASD

Rn 363 kips   1.67  217 kips  40 kips

o.k.

Flexural Yielding of the Plate The required flexural strength is calculated based upon the required shear strength and the eccentricity previously calculated: LRFD

M u  Vu e

ASD

M a  Va e

  75 kips 1.25 in.

  50 kips 1.25 in.

 93.8 kip-in.

 62.5 kip-in.

From AISC Manual Part 10, the plate buckling will not control for the conventional configuration. The flexural yielding strength is determined as follows: Zg  

t pl 2 4

2 in.14.5 in.2 4 3

 26.3 in.

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M n  Fy Z g



  50 ksi  26.3 in.3



 1,320 kip-in.

LRFD

  0.90

ASD

  1.67

M n  0.90 1,320 kip-in.

M n 1,320 kip-in.   1.67  790 kip-in.  62.5 kip-in.

1,190 kip-in.  93.8 kip-in. o.k.

o.k.

Interaction of Axial, Flexural and Shear Yielding in Plate AISC Specification Chapter H does not address combined flexure and shear. The method employed here is derived from Chapter H in conjunction with AISC Manual Equation 10-5. LRFD

ASD

Nu 60 kips  Rnp 327 kips

Na Rnp

 0.183

Because

 0.184

Nu  0.2 : Rnp

 Nu M  u   2Rnp M n

40 kips   217 kips

2

Because 2

  Vu      1   Rnv  2

Na  0.2 : Rnp 

 N a M a   Mn  2 Rnp 2

 60 kips 93.8 kip-in.   75 kips       1  2  327 kips  1,190 kip-in.   218 kips  0.147  1 o.k.

2

  Va 2     1   Rnv  2

2

 40 kips 62.5 kip-in.   50 kips       1  2  217 kips  790 kip-in.   145 kips  0.148  1 o.k.

Shear Rupture Strength of the Plate From AISC Specification Section J4.2(b), the available shear rupture strength of the plate is determined as follows: Anv  l  n  d h  z in.  t  14.5 in. – 5 , in.  z in.  2 in.  4.75 in.2

Rn  0.60 Fu Anv



 0.60  65 ksi  4.75 in.

2

(Spec. Eq. J4-4)



 185 kips

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LRFD

  0.75

Rn  0.75 185 kips   139 kips  75 kips

  2.00

Rn 185 kips   2.00  92.5 kips  50 kips

o.k.

ASD

o.k.

Tensile Rupture of the Plate From AISC Specification Section J4.1(b), the available tensile rupture strength of the plate is determined as follows: An  l  n  d h  z in.  t  14.5 in. – 5 , in.  z in.  2 in.  4.75 in.2

Table D3.1, Case 1, applies in this case because the tension load is transmitted directly to the cross-sectional element by fasteners; therefore, U = 1.0. Ae  AnU



(Spec. Eq. D3-1)



 4.75 in.2 1.0   4.75 in.2

Rn  Fu Ae



  65 ksi  4.75 in.

2

(Spec. Eq. J4-2)



 309 kips LRFD

  0.75

Rn  0.75  309 kips   232 kips  60 kips

  2.00

Rn 309 kips   2.00  155 kips  40 kips

o.k.

ASD

o.k.

Flexural Rupture of the Plate The available flexural rupture strength of the plate is determined as follows:





tp   d h  z in. s  n2  1   dh  z in.2  4 2 in.   26.3 in.3  , in.  z in. 3.00 in. 52  1  , in.  z in.2  4 

Z net  Z g 





 17.2 in.3 M n  Fu Z net

(Manual Eq. 9-4)



  65 ksi  17.2 in.

3



 1,120 kip-in.

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LRFD

  0.75

M n  0.75 1,120 kip-in.  840 kip-in.  93.8 kip-in.

ASD

  2.00

o.k.

M n 1,120 kip-in.   2.00  560 kip-in.  62.5 kip-in.

o.k.

Interaction of Axial, Flexure and Shear Rupture in Plate AISC Specification Chapter H does not address combined flexure and shear. The method employed here is derived from Chapter H in conjuction with AISC Manual Equation 10-5. LRFD

ASD Na

Nu 60 kips  Rnp 232 kips

Rnp

40 kips   155 kips  0.258

 0.259

Because

Nu  0.2 : Rnp

 Nu 8 M u    Rnp 9 M n

2

Because

2

Rnp 

 N a 8 M a    Rnp 9 M n

  Vu      1   Rnv  2

Na

2

 60 kips 8  93.8 kip-in.    75 kips         1  232 kips 9  840 kip-in.    139 kips  0.419  1 o.k.

 0.2 : 2

2

  Va      1   Rnv  2

2

 40 kips 8  62.5 kip-in.    50 kips         1 155 kips 9  560 kip-in.    92.5 kips  0.420  1 o.k.

Block Shear Rupture Strength of the Plate—Beam Shear Direction The nominal strength for the limit state of block shear rupture of the angles, assuming an L-shaped tearout due the shear load only, is determined as follows:

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant where Agv   l  lev  t  14.5 in.  14 in. 2 in.  6.63 in.2

Anv  Agv   n  0.5  d h  z in. t  6.63 in.2   5  0.5 , in.  z in.2 in.  4.38 in.2 Ant  leh  0.5  d h  z in.  t   22 in.  0.5 , in.  z in.  2 in.  1.00 in.2 U bs  1.0

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and















Rn  0.60  65 ksi  4.38 in.2  1.0  65 ksi  1.00 in.2  0.60  50 ksi  6.63 in.2  1.0  65 ksi  1.00 in.2



 236 kips  264 kips Therefore: Rn  236 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture on the plate is: LRFD

  0.75

Rn  0.75  236 kips   177 kips  75 kips

ASD

  2.00

Rn 236 kips   2.00  118 kips  50 kips

o.k.

o.k.

Block Shear Rupture Strength of the Plate—Beam Axial Direction The plate block shear rupture failure path due to axial load only could occur as an L- or U-shape. Assuming an Lshaped failure path due to axial load only, the available block shear rupture strength of the plate is:

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Agv  leh t   22 in.2 in.  1.25 in.2

Anv  Agv  0.5  d h  z in. t

 1.25 in.2 – 0.5 , in.  z in.2 in.  1.00 in.2

Ant  l  lev   n  0.5  d h  z in.  t  14.5 in.  14 in.   5  0.5 , in.  z in.  2 in.  4.38 in.2

U bs  1.0

and















Rn  0.60  65 ksi  1.00in.2  1.0  65 ksi  4.38 in.2  0.60  50 ksi  1.25 in.2  1.0  65 ksi  4.38 in.2



 324 kips  322 kips Therefore: Rn  322 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture on the plate is:

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LRFD

  0.75

Rn  0.75  322 kips   242 kips  60 kips

ASD

  2.00

Rn 322 kips   2.00  161 kips  40 kips

o.k.

o.k .

Assuming a U-shaped failure path in the plate due to axial load, the available block shear rupture strength of the plate is:

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Agv   2 shear planes  leh t p   2 shear planes  22 in.2 in.  2.50 in.2

Anv  Agv   2 shear planes  0.5  d h  z in. t

 2.50 in.2 –  2 shear planes  0.5 , in.  z in.2 in.  2.00 in.2

Ant  l  2lev   n  1 d h  z in.  t  14.5 in.  2 14 in.   5  1, in.  z in.  2 in.  4.00 in.2

U bs  1.0

and















Rn  0.60  65 ksi  2.00 in.2  1.0  65 ksi  4.00 in.2  0.60  50 ksi  2.50 in.2  1.0  65 ksi  4.00 in.2



 338 kips  335 kips Therefore: Rn  335 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture on the plate is:   0.75

LRFD

Rn  0.75  335 kips   251 kips  60 kips

  2.00

Rn 335 kips   2.00  168 kips  40 kips

o.k.

ASD

o.k .

The L-shaped failure path controls in the shear plate. Check shear and tension interaction for plate block shear on the L-shaped failure plane:

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LRFD 2

ASD 2

2

2

 Va   N a      1  Rnv    Rnt  

 Vu   Nu      1  Rnv   Rnt  2

2

 75 kips   60 kips       0.241  1 o.k.  177 kips   242 kips 

2

2

 50 kips   40 kips       0.241  1 o.k.  118 kips   161 kips 

Shear Strength of the Beam Web From AISC Specification Section J4.2(a), the available shear yielding strength of the beam is determined as follows: Agv  dtw  18.0 in. 0.355 in.  6.39 in.2 Rn  0.60 Fy Agv



 0.60  50 ksi  6.39 in.

2

(Spec. Eq. J4-3)



 192 kips

LRFD

  1.00

Rn  1.00 192 kips   192 kips  75 kips

o.k.

  1.50

Rn 192 kips   1.50  128 kips  50 kips

ASD

o.k.

The limit state of shear rupture of the beam web will not control in this example because the beam is uncoped. Tensile Strength of the Beam From AISC Specification Section J4.1(a), the available tensile yielding strength of the beam is determined as follows: Rn  Fy Ag   50 ksi  14.7 in.2  735 kips



  0.90

(Spec. Eq. J4-1)



LRFD

Rn  0.90  735 kips   662 kips  60 kips

o.k.

  1.67

Rn 735 kips   1.67  440 kips  40 kips

ASD

o.k.

From AISC Specification Sections J4.1, the available tensile rupture strength of the beam is determined from AISC Specification Equation J4-2. No cases in Table D3.1 apply to this configuration; therefore, U is determined in accordance with AISC Specification Section D3, where the minimum value of U is the ratio of the gross area of the connected element to the member gross area.

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U 

 d  2t f  tw Ag 18.0 in.  2  0.570 in.   0.355 in. 14.7 in.2

 0.407

An  Ag  n  d h  z in. t w

 14.7 in.2  5 , in.  z in. 0.355 in.  12.9 in.

Ae  AnU



 12.9 in.2

(Spec. Eq. D3-1)

  0.407 

 5.25 in.2 Rn  Fu Ae



  65 ksi  5.25 in.

2

(Spec. Eq. J4-2)



 341 kips

  0.75

LRFD

Rn  0.75  341 kips   256 kips  60 kips

  2.00

Rn 341 kips   2.00  171 kips  40 kips

o.k.

ASD

o.k.

Block Shear Rupture of the Beam Web Block shear rupture is only applicable in the direction of the axial load, because the beam is uncoped and the limit state is not applicable for an uncoped beam subject to vertical shear. Assuming a U-shaped tearout relative to the axial load, and assuming a horizontal edge distance of leh = 1w in.  4 in. = 12 in. to account for a possible beam underrun of 4 in., the block shear rupture strength is:

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant where Agv   2 shear planes  leh tw   2 shear planes 12 in. 0.355 in.  1.07 in.2

Anv  Agv   2 shear planes  0.5  d h  z in. tw  1.07 in.2   2 shear planes  0.5, in.  z in. 0.355 in.  0.715 in.2

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Ant  12.0 in.   n  1 dh  z in.  tw  12.0 in.   5  1, in.  z in.   0.355 in.  2.84 in.2 U bs  1.0

and















Rn  0.60  65 ksi  0.715 in.2  1.0  65 ksi  2.84 in.2  0.60  50 ksi  1.07 in.2  1.0  65 ksi  2.84 in.2



 212 kips  217 kips Therefore: Rn  212 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture of the beam web is:

Rn  0.75  212 kips 

LRFD

 159 kips  60 kips

o.k.

ASD Rn 212 kips   2.00  106 kips  40 kips

o.k.

Conclusion The connection is found to be adequate as given for the applied loads. Note that because the supported member was assumed to be continuously laterally braced, it is not necessary to check weak-axis moment.

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EXAMPLE II.A-17C SINGLE-PLATE CONNECTION—STRUCTURAL INTEGRITY CHECK Given: Verify the single plate connection from Example II.A-17A, as shown in Figure II.A-17C-1, for the structural integrity provisions of AISC Specification Section B3.9. The connection is verified as a beam and girder end connection and as an end connection of a member bracing a column. Note that these checks are necessary when design for structural integrity is required by the applicable building code. Use 70-ksi electrodes and an ASTM A36 plate.

Fig. II.A-17C-1. Connection geometry for Example II.A-17C. Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Beam W16x50

tw = 0.380 in. From AISC Specification Table J3.3, the hole diameter for w-in.-diameter bolts with standard holes is: dh = m in. Beam and Girder End Connection From Example II.A-17A, the required shear strength is: LRFD

ASD

Vu  49.6 kips

Va  33.0 kips

From AISC Specification Section B3.9(b), the required axial tensile strength is: LRFD 2 Tu  Vu  10 kips 3 2   49.6 kips   10 kips 3  33.1 kips  10 kips  33.1 kips

ASD Ta  Va  10 kips  33.0 kips  10 kips  33.0 kips

Bolt Shear From AISC Specification Section J3.6, the nominal bolt shear strength is: Fnv = 54 ksi, from AISC Specification Table J3.2

Tn  nFnv Ab



  4 bolts  54 ksi  0.442 in.

2

(from Spec. Eq. J3-1)



 95.5 kips Bolt Bearing and Tearout From AISC Specification Section B3.9, for the purpose of satisfying structural integrity requirements inelastic deformations of the connection are permitted; therefore, AISC Specification Equations J3-6b and J3-6d are used to determine the nominal bearing and tearout strength. By inspection, bolt bearing and tearout will control for the plate. For bolt bearing on the plate: Tn   4 bolts  3.0dtFu

(from Spec. Eq. J3-6b)

  4 bolts  3.0  w in.4 in. 58 ksi   131 kips

For bolt tearout on the plate:

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lc  leh  0.5  d h  z in.  12 in.  0.5 m in.  z in.  1.06 in. Tn   4 bolts 1.5lc tFu

(from Spec. Eq. J3-6d)

  4 bolts 1.5 1.06 in.4 in. 58 ksi   92.2 kips

Tensile Yielding of Plate From AISC Specification Section J4.1, the nominal tensile yielding strength of the shear plate is determined as follows: Ag  lt  11.5 in.4 in.  2.88 in.2 Tn  Fy Ag

(from Spec. Eq. J4-1)



  36 ksi  2.88 in.2



 104 kips

Tensile Rupture of Plate From AISC Specification Section J4.1, the nominal tensile rupture strength of the shear plate is determined as follows: An  l  n  d h  z in.  t  11.5 in.   4 bolts m in.  z in.  4 in.  2.00 in.2

AISC Specification Table D3.1, Case 1 applies in this case because tension load is transmitted directly to the crosssection element by fasteners; therefore, U = 1.0. Ae  AnU



2

 2.00 in.

(Spec. Eq. D3-1)

 1.0

 2.00 in.2 Tn  Fu Ae

(from Spec. Eq. J4-2)



  58 ksi  2.00 in.2



 116 kips

Block Shear Rupture—Plate From AISC Specification Section J4.3, the nominal block shear rupture strength, due to axial load, of the shear plate is determined as follows:

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Tn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(from Spec. Eq. J4-5)

where Agv   2 shear planes  leh t   2 shear planes  12 in.4 in.  0.750 in.2 Anv   2 shear planes  leh  0.5  d h  z in.  t p   2 shear planes  12 in.  0.5 m in.  z in.  4 in.  0.531 in.2

Ant  l  2lev   n  1 d h  z in.  t  11.5 in.  2 14 in.   4  1m in.  z in.  4 in.  1.59 in.2

U bs  1.0

and Tn  0.60  58 ksi  0.531 in.2  1.0  58 ksi  1.59 in.2  0.60  36 ksi  0.750 in.2  1.0  58 ksi  1.59 in.2

















 111 kips  108 kips  108 kips

Block Shear Rupture—Beam Web From AISC Specification Section J4.3, the nominal block shear rupture strength, due to axial load, of the beam web is determined as follows (accounting for a possible 4-in. beam underrun): Tn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

where Agv   2 shear planes  leh  underrun  tw   2 shear planes  22 in.  4 in. 0.380 in.  1.71 in.2 Anv   2 shear planes  leh  underrun  0.5  d h  z in.  t w   2 shear planes   22 in.  4 in.  0.5 m in.  z in.   0.380 in.  1.38 in.2

Ant  9.00 in.  3 m in.  z in.   0.380 in.

 2.42 in.2 U bs  1.0

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and Tn  0.60  65 ksi  1.38 in.2  1.0  65 ksi  2.42 in.2  0.60  50 ksi  1.71 in.2  1.0  65 ksi  2.42 in.2

















 211 kips  209 kips  209 kips

Weld Strength From AISC Specification Section J2.4, the nominal tensile strength of the weld is determined as follows:



Fnw  0.60 FEXX 1.0  0.50sin1.5 





(Spec. Eq. J2-5)

 0.60  70 ksi  1.0  0.50sin1.5 90



 63.0 ksi

The throat dimension is used to calculate the effective area of the fillet weld. w

l  2 welds  2 x in.  11.5 in. 2 welds  2

Awe 

 3.05 in.2 Tn  Fnw Awe

(from Spec. Eq. J2-4)



  63.0 ksi  3.05 in.

2



 192 kips

Nominal Tensile Strength The controlling tensile strength, Tn, is the least of those previously calculated:

Tn  min 95.5 kips, 131 kips, 92.2 kips, 104 kips, 116 kips, 108 kips, 209 kips, 192 kips  92.2 kips LRFD Tn  92.2 kips  33.1 kips o.k.  

ASD Tn  92.2 kips  33.0 kips o.k.  

Column Bracing From AISC Specification Section B3.9(c), the minimum axial tension strength for the connection of a member bracing a column is equal to 1% of two-thirds of the required column axial strength for LRFD and equal to 1% of the required column axial for ASD. These requirements are evaluated independently from other strength requirements. The maximum column axial force this connection is able to brace is determined as follows: LRFD

2  Tn  0.01 Pu    3 

ASD Tn  0.01Pa  

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LRFD Solving for the column axial force:

ASD Solving for the column axial force:

3  Pu  100  Tn  2  3  100    92.2 kips    2  13,800 kips

Pa  100Tn  100  92.2 kips     9, 220 kips

As long as the required column axial strength is less than Pu = 13,800 kips or Pa = 9,220 kips, this connection is an adequate column brace.

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EXAMPLE II.A-18 SINGLE-PLATE CONNECTION (BEAM-TO-GIRDER WEB) Given: Verify a single-plate connection between an ASTM A992 W1835 beam and an ASTM A992 W2162 girder web, as shown in Figure II.A-18-1, to support the following beam end reactions: RD = 6.5 kips RL = 20 kips The top flange is coped 2 in. deep by 4 in. long, lev = 12 in. Use 70-ksi electrodes and an ASTM A36 plate.

Fig. II.A-18-1. Connection geometry for Example II.A-18.

Solution: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam and girder ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Beam W1835 tw = 0.300 in. d = 17.7 in. tf = 0.425 in. Girder W2162 tw = 0.400 in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  6.5 kips   1.6  20 kips 

ASD

Ra  6.5 kips  20 kips  26.5 kips

 39.8 kips Connection Selection

AISC Manual Table 10-10a includes checks for the limit states of bolt shear, bolt bearing on the plate, tearout on the plate, shear yielding of the plate, shear rupture of the plate, block shear rupture of the plate and weld shear. Use four rows of bolts, 4-in. plate thickness, and x-in. fillet weld size. From AISC Manual Table 10-10a: LRFD

ASD Rn  34.8 kips  26.5 kips o.k. 

Rn  52.2 kips  39.8 kips o.k.  Block Shear Rupture of Beam Web

The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the beam web is determined as follows, using AISC Manual Tables 9-3a, 9-3b and 9-3c and AISC Specification Equation J4-5, with n = 4, leh = 24 in. (reduced 4 in. to account for beam underrun), lev = 12 in. and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a: 

Fu Ant  88.4 kip/in.  t





Shear yielding component from AISC Manual Table 9-3b: 

0.60Fy Agv  236 kip/in.   t

ASD Tension rupture component from AISC Manual Table 9-3a:

Fu Ant  58.9 kip/in. t

Shear yielding component from AISC Manual Table 9-3b:





0.60Fy Agv  158 kip/in.  t

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LRFD Shear rupture component from AISC Manual Table 9-3c:



0.60Fu Anv  218 kip/in.   t  The design block shear rupture strength is: Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant



  218 kip/in.  88.4 kip/in. 0.300 in.   236 kip/in.  88.4 kip/in. 0.300 in.  91.9 kips  97.3 kips

ASD Shear rupture component from AISC Manual Table 9-3c:





0.60Fu Anv  145 kip/in. t

The allowable block shear rupture strength is: Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant  +    145 kip/in.  58.9 kip/in. 0.300 in.  158 kip/in.  58.9 kip/in. 0.300 in.  61.2 kips  65.1 kips

Therefore:

Therefore:

Rn  91.9 kips  39.8 kips

o.k.

Rn  61.2 kips  26.5 kips o.k. 

Strength of the Bolted Connection—Beam Web From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the individual strengths of the individual fasteners, which may be taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is: LRFD

ASD rn  11.9 kips/bolt 

rn  17.9 kips/bolt

The available bearing and tearout strength of the beam web edge bolt (top bolt shown in Figure II.A-18-1) is determined using AISC Manual Table 7-5, conservatively using le = 14 in. LRFD

rn   49.4 kip/in. 0.300 in.  14.8 kips/bolt

ASD rn   32.9 kip/in. 0.300 in.   9.87 kips/bolt

The bearing or tearout strength controls over bolt shear for the edge bolt. The available bearing and tearout strength of the beam web at the interior bolts is determined using AISC Manual Table 7-4 with s = 3 in.

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LRFD

ASD rn   58.5 kip/in. 0.300 in.   17.6 kips/bolt

rn   87.8 kip/in. 0.300 in.  26.3 kips/bolt Bolt shear strength controls for the interior bolts.

The strength of the bolt group in the beam web is determined by summing the strength of the individual fasteners as follows: LRFD

ASD

Rn

Rn  1 bolt 14.8 kips/bolt 



  3 bolts 17.9 kips/bolt 

 68.5 kips  39.8 kips o.k.

 1 bolt  9.87 kips/bolt    3 bolts 11.9 kips/bolt 

 45.6 kips  26.5 kips o.k.

Strength of the Bolted Connection—Single Plate The available bearing and tearout strength of the plate at the edge bolt (bottom bolt shown in Figure II.A-18-1) is determined using AISC Manual Table 7-5 with le = 14 in. LRFD

ASD rn   29.4 kip/in.4 in.   7.35 kips/bolt

rn   44.0 kip/in.4 in.  11.0 kips/bolt

The bearing or tearout strength controls over bolt shear for the edge bolt. The available bearing and tearout strength of the plate at the interior bolts is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

ASD rn   52.2 kip/in.4 in.   13.1 kips/bolt

rn   78.3 kip/in.4 in.  19.6 kips/bolt Bolt shear strength controls for the interior bolts.

The strength of the bolt group in the plate is determined by summing the strength of the individual fasteners as follows: LRFD

ASD

Rn

Rn  1 bolt 11.0 kips/bolt 



  3 bolts 17.9 kips/bolt   64.7 kips  39.8 kips o.k.

 1 bolt  7.35 kips/bolt    3 bolts 11.9 kips/bolt   43.1 kips  26.5 kips o.k.

Shear Rupture of the Girder Web at the Weld The minimum support thickness to match the shear rupture strength of the weld is determined as follows:

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tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  3 sixteenths 

65 ksi  0.143 in.  0.400 in. o.k. Note: For coped beam sections, the limit states of flexural yielding and local buckling should be checked independently per AISC Manual Part 9. The supported beam web should also be checked for shear yielding and shear rupture per AISC Specification Section J4.2. However, for the shallow cope in this example, these limit states do not govern. For an illustration of these checks, see Example II.A-4. Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-19A

EXTENDED SINGLE-PLATE CONNECTION (BEAM-TO-COLUMN WEB)

Given: Verify the connection between an ASTM A992 W1636 beam and the web of an ASTM A992 W1490 column, as shown in Figure II.A-19A-1, to support the following beam end reactions: RD = 6 kips RL = 18 kips Use 70-ksi electrodes and ASTM A36 plate.

Fig. II.A-19A-1. Connection geometry for Example II.A-19A. Note: All dimensional limitations are satisfied.

Solution: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Beam W1636 tw = 0.295 in. d = 15.9 in. Column W1490 tw = 0.440 in. bf = 14.5 in. From AISC Specification Table J3.3, the hole diameter for a w-in.-diameter bolt with standard holes is: d h  m in.

From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  6 kips   1.6 18 kips 

ASD

Ra  6 kips  18 kips  24.0 kips

 36.0 kips Strength of the Bolted Connection—Beam Web

From AISC Manual Part 10, determine the distance from the support to the first line of bolts and the distance to the center of gravity of the bolt group. a  9 in. 3in. 2  10.5 in.

e  9 in. 

From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is: LRFD

ASD rn  11.9 kips 

rn  17.9 kips

Tearout for the bolts in the beam web does not control due to the presence of the beam top flange. The available bearing strength of the beam web per bolt is determined using AISC Manual Table 7-4 with s = 3 in. LRFD rn   87.8 kip/in. 0.295 in.

ASD rn =  58.5 kip/in. 0.295 in.   17.3 kips

 25.9 kips Therefore, bolt shear controls over bearing.

The strength of the bolt group is determined by interpolating AISC Manual Table 7-7, with e = 10.5 in. and Angle = 0: C = 2.33

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LRFD

ASD

Rn  C rn

Rn Crn     2.33 11.9 kips 

 2.33 17.9 kips   41.7 kips > 36.0 kips

o.k.

= 27.7 kips > 24.0 kips o.k. Maximum Plate Thickness From AISC Manual Part 10, determine the maximum plate thickness, tmax, that will result in the plate yielding before the bolts shear. Fnv = 54 ksi from AISC Specification Table J3.2

C  = 26.0 in. from AISC Manual Table 7-7 for the moment-only case (Angle = 0) Fnv  Ab C'  0.90  54 ksi  2   0.442 in.  0.90   690 kip-in.

M max 



tmax = 

(Manual Eq. 10-4)

  26.0 in.

6M max

(Manual Eq. 10-3)

Fy l 2 6  690 kip-in.

 36 ksi 12.0 in.2

 0.799 in. Try a plate thickness of 2 in. Strength of the Bolted Connection—Plate The available bearing strength of the plate per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration, as follows:

rn  2.4dtFu

(Spec. Eq. J3-6a)

 2.4  w in.2 in. 58 ksi   52.2 kips/bolt   0.75

LRFD

rn  0.75  52.2 kips/bolt   39.2 kips/bolt

  2.00

ASD

rn 52.2 kips/bolt =  2.00  26.1 kips/bolt

The available tearout strength of the bottom edge bolt in the plate is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration, as follows:

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lc  leh  0.5d h

 12 in.  0.5 m in.  1.09 in.

rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2 1.09 in.2 in. 58 ksi   37.9 kips/bolt LRFD

  0.75

  2.00

rn  0.75  37.9 kips/bolt 

ASD

rn 37.9 kips/bolt =  2.00  19.0 kips/bolt

 28.4 kips/bolt

Therefore, the bolt shear determined previously controls for the bolt group in the plate. Shear Strength of Plate From AISC Specification Section J4.2(a), the available shear yielding strength of the plate is determined as follows: Agv  lt  12.0 in.2 in.  6.00 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  6.00 in.2



 130 kips

LRFD

  1.00

  1.50

Rn  1.00 130 kips 

ASD

Rn 130 kips   1.50  86.7 kips  24.0 kips o.k.

 130 kips  36.0 kips o.k.

From AISC Specification Section J4.2(b), the available shear rupture strength of the plate is determined using the net area determined in accordance with AISC Specification Section B4.3b.

Anv  l  n  d h  z in.  t  12.0 in.  4 m in.  z in.  2 in.  4.25 in.2

Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  58 ksi  4.25 in.2



 148 kips

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LRFD

  0.75

ASD

  2.00

Rn  0.75 148 kips 

Rn 148 kips  2.00   74.0 kips  24.0 kips o.k.

 111 kips  36.0 kips o.k. Block Shear Rupture of Plate

From AISC Specification Section J4.3, the block shear rupture strength of the plate is determined as follows.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Agv   l  lev  t  12.0 in.  12 in.2 in.  5.25 in.2

Anv  Agv   n  0.5  d h  z in. t

 5.25 in.2   4  0.5 m in.  z in.2 in.  3.72 in.2 Ant  3 in.  14 in.  1.5  d h  z in.  t  3 in.  14 in.  1.5 m in.  z in.  2 in.  1.47 in.2 U bs  0.5

and















Rn  0.60  58 ksi  3.72 in.2  0.5  58 ksi  1.47 in.2  0.60  36 ksi  5.25 in.2  0.5  58 ksi  1.47 in.2



 172 kips  156 kips Therefore:

Rn  156 kips   0.75

LRFD

  2.00

Rn  0.75 156 kips 

ASD

Rn 156 kips  2.00   78.0 kips  24.0 kips

 117 kips  36.0 kips o.k.

o.k.

Interaction of Shear Yielding and Flexural Yielding of Plate From AISC Manual Part 10, the plate is checked for the interaction of shear yielding and yielding due to flexure as follows:

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LRFD 2

ASD

2

2

 Vr   M r   V    M   1.0  c  c

(Manual Eq. 10-5)

2

 Vr   M r   V    M   1.0  c  c

(Manual Eq. 10-5)

From the preceding calculations:

From the preceding calculations:

Vr  Vu  36.0 kips

Vr  Va  24.0 kips

Vc  vVn  130 kips

Vc 

From AISC Manual Part 10:

From AISC Manual Part 10:

Vn v  86.7 kips

M c  b M n

Mn b Fy Z pl  b

Mc 

 b Fy Z pl  2 in.12 in.2    0.90  36 ksi   4    583 kip-in.

2  36 ksi   2 in.12 in.      4  1.67   

 388 kip-in.

Mr  Mu

Mr  Ma

 Vu a

 Va a

  36.0 kips  9 in.

  24.0 kips  9 in.

 324 kip-in.

 216 kip-in.

2

2

 36.0 kips   324 kip-in.       0.386  1.0  130 kips   583 kip-in. 

2

o.k.

2

 24.0 kips   216 kip-in.       0.387  1.0  86.7 kips   388 kip-in. 

o.k.

Lateral-Torsional Buckling of Plate The plate is checked for the limit state of buckling using the double-coped beam procedure as given in AISC Manual Part 9, where the unbraced length for lateral-torsional buckling, Lb, is taken as the distance from the first column of bolts to the supporting column web and the top cope dimension, dct, is conservatively taken as the distance from the top of the beam to the first row of bolts.  L Cb  3  ln  b  d 

   d ct   1  d  

   1.84 

(Manual Eq. 9-15)

3 in.    9 in.     3  ln    1    1.84  12 in.    12 in.    2.03  1.84 Therefore:

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Cb  2.03

From AISC Specification Section F11, the flexural strength of the plate for the limit state of lateral-torsional buckling is determined as follows: Lb d t

2



 9 in.12 in. 2 in.2

 432

0.08E 0.08  29, 000 ksi   36 ksi Fy  64.4 1.9 E 1.9  29, 000 ksi   Fy 36 ksi  1,530

Because

0.08E Lb d 1.9E  2  , use AISC Specification Section F11.2(b): Fy Fy t

M p  Fy Z x  2 in.12 in.2     36 ksi   4    648 kip-in. M y  Fy S x  2 in.12 in.2     36 ksi   6    432 kip-in.

  L d  Fy  M n  Cb 1.52  0.274  b2   M y  M p  t  E    36 ksi     2.03 1.52  0.274  432      432 kip-in.  648 kip-in.  29,000 ksi     1,200 kip-in.  648 kip-in. Therefore: M n  648 kip-in.

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LRFD

ASD

b  0.90

 b  1.67

b M n  0.90  648 kip-in.

M n 648 kip-in  b 1.67  388 kip-in.  216 kip-in.

 583 kip-in.  324 kip-in. o.k.

o.k.

Flexural Rupture of Plate The net plastic section modulus of the plate, Znet, is determined from AISC Manual Table 15-3:

Z net  12.8 in.3 From AISC Manual Equation 9-4: M n  Fu Z net

(Manual Eq. 9-4)



  58 ksi  12.8 in.3



 742 kip-in.

LRFD

ASD

b  0.75

 b  2.00 

b M n  0.75  742 kip-in.

M n 742 kip-in.   2.00  371 kip-in. > 216 kip-in.

 557 kip-in. > 324 kip-in. o.k.

o.k.

Weld Between Plate and Column Web (AISC Manual Part 10) From AISC Manual Part 10, a weld size of (s)tp is used to develop the strength of the shear plate. w  st p

 s 2 in.  c in. Use a two-sided c-in. fillet weld. Strength of Column Web at Weld The minimum column web thickness to match the shear rupture strength of the weld is determined as follows:

tmin = 

3.09D Fu

(Manual Eq. 9-2)

3.09  5 sixteenths 

65 ksi  0.238 in.  0.440 in. o.k. Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.A-19B EXTENDED SINGLE-PLATE CONNECTION SUBJECT TO AXIAL AND SHEAR LOADING Given: Verify the available strength of an extended single-plate connection for an ASTM A992 W1860 beam to the web of an ASTM A992 W1490 column, as shown in Figure II.A-19B-1, to support the following beam end reactions: LRFD Shear, Vu = 75 kips Axial tension, Nu = 60 kips

ASD Shear, Va = 50 kips Axial tension, Na = 40 kips

Use 70-ksi electrodes and ASTM A572 Grade 50 plate.

Fig. II.A-19B-1. Connection geometry for Example II.A-19B. Solution: From AISC Manual Table 2-4 and Table 2-5, the material properties are as follows: Beam, column ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A572 Grade 50 Fy = 50 ksi Fu = 65 ksi

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From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1860

Ag d tw bf tf

= 17.6 in.2 = 18.2 in. = 0.415 in. = 7.56 in. = 0.695 in.

Column W1490

d = 14.0 in. tw = 0.440 in. kdes = 1.31 in. From AISC Specification Table J3.3, for 1-in.-diameter bolts with standard holes: dh = 18 in. Per AISC Specification Section J3.2, standard holes are required for both the plate and beam web because the beam axial force acts longitudinally to the direction of a slotted hole and bolts are designed for bearing. The resultant load is determined as follows: LRFD 2

Ru  Vu  N u 

ASD

2

2

Ra  Va  N a

 75 kips 2   60 kips 2



 96.0 kips

2

 50 kips 2   40 kips 2

 64.0 kips

The resultant load angle is determined as follows: LRFD

ASD

 60 kips    tan 1    75 kips   38.7

 40 kips    tan 1    50 kips   38.7

Strength of Bolted Connection—Beam Web The strength of the bolt group is determined by interpolating AISC Manual Table 7-7 for Angle  30 and n = 5. Note that 300 is used conservatively in order to employ AISC Manual Table 7-7. A direct analysis can be performed to obtain an accurate value using the instantaneous center of rotation method. ex  a  0.5s  9w in.  0.5  3 in.  11.3 in.

C  3.53 by interpolation

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From AISC Manual Table 7-1, the available shear strength per bolt for 1-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is: LRFD

ASD

rn  21.2 kips/bolt 

rn  31.8 kips/bolt

The available bearing strength of the beam web is determined from AISC Specification Equation J3-6b. This equation is applicable in lieu of Equation J3-6a, because plowing of the bolts in the beam web is desirable to provide some flexibility in the connection:

rn  3.0dtw Fu  3.0 1 in. 0.415 in. 65 ksi 

(Spec. Eq. J3-6b)

 80.9 kips/bolt   0.75

LRFD

  2.00

rn  0.75  80.9 kips/bolt 

ASD

rn 80.9 kips/bolt   2.00  40.5 kips/bolt

 60.7 kips/bolt

The available tearout strength of the beam web is determined from Specification Equation J3-6d. Similar to the bearing strength determination, this equation is used to allow plowing of the bolts in the beam web, and thus provide some flexibility in the connection. Because the direction of load on the bolt is unknown, the minimum bolt edge distance is used to determine a worst case available tearout strength (including a 4-in. tolerance to account for possible beam underrun). If a computer program is available, the true le can be calculated based on the instantaneous center of rotation. lc  leh  0.5d h  1w in.  4 in.  0.5 18 in.  0.938 in. rn  1.5lc t w Fu

(Spec. Eq. J3-6d)

 1.5  0.938 in. 0.415 in. 65 ksi   38.0 kips/bolt

  0.75

LRFD

  2.00

rn  0.75  38.0 kips/bolt 

ASD

rn 38.0 kips/bolt   2.00  19.0 kips/bolt

 28.5 kips/bolt

The tearout strength controls for bolts in the beam web. The available strength of the bolted connection is determined using the minimum available strength calculated for bolt shear, bearing on the beam web and tearout on the beam web. From AISC Manual Equation 7-16, the bolt group eccentricity is accounted for by multiplying the minimum available bolt strength by the bolt coefficient C.

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LRFD

Rn  C rn  3.53  28.5 kips/bolt   101 kips > 96.0 kips

o.k.

ASD Rn rn C    3.53 19.0 kips/bolt 

 67.1 kips > 64.0 kips

o.k.

Strength of Bolted Connection—Plate Note that bolt bearing on the beam web controls over bearing on the plate because the beam web is thinner than the plate; therefore, this limit state will not control. As was discussed for the beam web, the available tearout strength of the plate is determined from Specification Equation J3-6d. The bolt edge distance in the vertical direction controls for this design. lc  lev  0.5d h  14 in.  0.5 18 in.  0.688 in. rn  1.5lc tFu

(Spec. Eq. J3-6d)

 1.5  0.688 in. w in. 65 ksi   50.3 kips/bolt

LRFD

  0.75

rn  0.75  50.3 kips/bolt 

  2.00

ASD

rn 50.3 kips/bolt   2.00  25.2 kips/bolt

 37.7 kips/bolt

Therefore, the available strength of the bolted connection at the beam web, as determined previously, controls. Shear Yielding Strength of Beam From AISC Specification Section J4.2(a), the available shear yielding strength of the beam is determined as follows:

Agv  dtw  18.2 in. 0.415 in.  7.55 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  7.55 in.2



 227 kips

  1.00

LRFD

Rn  1.00  227 kips   227 kips  75 kips

o.k .

  1.50

Rn 227 kips   1.50  151 kips  50 kips

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Tensile Yielding Strength of Beam From AISC Specification Section J4.1(a), the available tensile yielding strength of the beam web is determined as follows: Rn  Fy Ag

(Spec. Eq. J4-1)



  50 ksi  17.6 in.2



 880 kips

LRFD

  0.90

Rn  0.90  880 kips   792 kips  60 kips

  1.67

Rn 880 kips   1.67  527 kips  40 kips

o.k .

ASD

o.k.

Tensile Rupture Strength of Beam From AISC Specification Section J4.1, determine the available tensile rupture strength of the beam. The effective net area is Ae = AnU, where U is determined from AISC Specification Table D3.1, Case 2. x 

2b f 2t f  tw2  d  2t f 8b f t f

  4tw  d  2t f 

2  7.56 in.  0.695 in.   0.415 in. 18.2 in.  2  0.695 in.   8  7.56 in. 0.695 in.  4  0.415 in. 18.2 in.  2  0.695 in.  2

2

 1.18 in.

x l 1.18 in.  1 3.00 in.  0.607

U  1

An  Ag  n  d h  z in. tw  17.6 in.2  5 18 in.  z in. 0.415 in.  15.1 in.2

Rn  Fu Ae  Fu AnU

(Spec. Eq. J4-2)





  65 ksi  15.1 in.2  0.607   596 kips

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LRFD

  0.75

Rn  0.75  596 kips   447 kips  60 kips

ASD

  2.00

Rn 596 kips   2.00  298 kips  40 kips

o.k.

o.k.

Block Shear Rupture of Beam Web Block shear rupture is only applicable in the direction of the axial load because the beam is uncoped and the limit state is not applicable for an uncoped beam subject to vertical shear. Assuming a U-shaped tearout relative to the axial load, and assuming a horizontal edge distance of leh = 1w in.  4 in. = 12 in. to account for a possible beam underrun of 4 in., the block shear rupture strength is:

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Agv   2 shear planes  s  leh  tw

  2 shear planes  3 in.  12 in. 0.415 in.  3.74 in.2

Anv  Agv   2 shear planes 1.5 d h  z in. tw  3.74 in.2   2 shear planes 1.5 18 in.  z in. 0.415 in.  2.26 in.2 Ant  12.0 in.   n  1 d h  z in.  tw  12.0 in.   5  118 in.  z in.   0.415 in.  3.01 in.2 U bs  1.0

and















Rn  0.60  65 ksi  2.26 in.2  1.0  65 ksi  3.01 in.2  0.60  50 ksi  3.74 in.2  1.0  65 ksi  3.01 in.2



 284 kips  308 kips

Therefore: Rn  284 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture of the beam web is:   0.75

LRFD

Rn  0.75  284 kips   213 kips  60 kips

o.k.

  2.00

Rn 284 kips   2.00  142 kips  40 kips

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Maximum Plate Thickness Determine the maximum plate thickness, tmax, that will result in the plate yielding before the bolts shear. From AISC Specification Table J3.2: Fnv = 54 ksi From AISC Manual Table 7-7 for two column of bolts, Angle = 0, s = 3 in., and n = 5: C   38.7 in.

Fnv  Ab C ' 0.90 54 ksi 0.785 in.2  38.7 in.  0.90  1,820 kip-in.

M max 



tmax  

(Manual Eq. 10-4)



6 M max

(Manual Eq. 10-3)

Fy l 2 6 1,820 kip-in.

 50 ksi 142 in.2

 1.04 in.  w in.

o.k.

Flexure Strength of Plate The required flexural strength of the plate is determined as follows: LRFD

ASD

M u  Vu a

M a  Va a

  75 kips  9w in.

  50 kips  9w in.

 731 kip-in.

 488 kip-in.

The plate is checked for the limit state of buckling using the double-coped beam procedure as given in AISC Manual Part 9, where the unbraced length for lateral-torsional buckling, Lb, is taken as the distance from the first column of bolts to the supporting column web and the top cope dimension, dct, is conservatively taken as the distance from the top of the beam to the first row of bolts.   L   d  Cb  3  ln  b    1  ct   1.84 d   d     38 in.   9w in.     3  ln    1    1.84  142 in.    142 in.    2.04  1.84

Therefore: Cb  2.04

The available flexural strength of the plate is determined using AISC Specification Section F11 as follows:

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For yielding of the plate: M n  M p  Fy Z  1.6 Fy S x

(Spec. Eq. F11-1)

  w in.142 in.    w in.142 in.    1.6  50 ksi      50 ksi   4 6      1,970 kip-in.  2,100 kip-in.  1,970 kip-in. 2

2

For lateral-torsional buckling of the plate: Lb d t

2



 9w in.142 in.  w in.2

 251

0.08 E 0.08  29, 000 ksi   Fy 50 ksi  46.4 1.9 E 1.9  29, 000 ksi   50 ksi Fy  1,100

Because

0.08E Lb d 1.9 E , use AISC Specification Section F11.2(b):  2  Fy Fy t

M y  Fy S x   w in.142 in.2     50 ksi   6    1,310 kip-in.   L d  Fy  M n  Cb 1.52  0.274  b2   M y  M p  t  E    50 ksi    2.04 1.52  0.274  251    1,310 kip-in.  1, 970 kip-in.  29, 000 ksi     3,750 kip-in.  1, 970 kip-in.

Therefore: M n  1,970 kip-in.

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(Spec. Eq. F11-2)

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LRFD

ASD

b  0.90

 b  1.67

b M n  0.90 1,970 kip-in.

M n 1,970 kip-in.  b 1.67  1,180 kip-in.  488 kip-in. o.k.

 1, 770 kip-in.  731 kip-in. o.k.

Shear Yielding Strength of Plate From AISC Specification Section J4.2(a), the available shear yielding strength of the plate is determined as follows: Agv  lt  142 in. w in.  10.9 in.2 Rnv  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  10.9 in.

2



 327 kips

  1.00

LRFD

  1.50

Rnv  1.00  327 kips   327 kips  75 kips

ASD

Rnv 327 kips   1.50  218 kips  50 kips

o.k.

o.k.

Tension Yielding Strength of Plate From AISC Specification Section J4.1(a), the available tensile yielding strength of the plate is determined as follows: Ag  lt  142 in. w in.  10.9 in.2 Rnp  Fy Ag

(from Spec. Eq. J4-1)

  50 ksi 10.9 in.  545 kips

  0.90

LRFD

  1.67

Rnp  0.90  545 kips   491 kips  60 kips

Rnp 

o.k.

ASD

545 kips 1.67  326 kips  40 kips 

Interaction of Axial, Flexure and Shear Yielding in Plate

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AISC Specification Chapter H does not address combined flexure and shear. The method employed here is derived from Chapter H in conjunction with AISC Manual Equation 10-5, as follows: LRFD

ASD N a 40 kips  Rnp 326 kips

Nu 60 kips  Rnp 491 kips

 0.123

 0.122

Because

Nu  0.2 : Rnp

 Nu Va  u   2Rnp M n

Because

2

N a  0.2 : Rnp 2

2

  Vu      1   Rnv 

2

 N a Va a   Va   1    M n   Rnv   2 Rnp

 60 kips  75 kips  9w in.     1, 770 kip-in.   2  491 kips 

2

2

 40 kips  50 kips  9w in.     1,180 kip-in.   2  326 kips 

2

2

 50 kips    1  218 kips   0.278  1 o.k.

 75 kips    1  327 kips   0.278  1 o.k.

Tensile Rupture Strength of Plate From AISC Specification Section J4.1(b), the available tensile rupture strength of the plate is determined as follows: An  l  n  d h  z in.  t  142 in. –  5 bolts 18 in.  z in.   w in.  6.42 in.2

AISC Specification Table D3.1, Case 1, applies in this case because the tension load is transmitted directly to the cross-sectional element by fasteners; therefore, U = 1.0. Ae  AnU



(Spec. Eq. D3-1)



 6.42 in.2 1.0   6.42 in.2

Rnp  Fu Ae



  65 ksi  6.42 in.

2

(Spec. Eq. J4-2)



 417 kips

  0.75

LRFD

Rnp  0.75  417 kips   313 kips  60 kips

o.k.

  2.00

ASD

Rnp 417 kips  2.00   209 kips  40 kips

Flexural Rupture of the Plate The available flexural rupture strength of the plate is determined as follows:

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Z net  

tl 2 t  2   d h  z in. s  n 2  1   d h  z in.   4 4



 w in.142 in.2 4





2  w in.   2    18 in.  z in. 3 in.  5   1  18 in.  z in.  4 



 23.1 in.3 M n  Fu Z net



  65 ksi  23.1 in.3

(Manual Eq. 9-4)



 1,500 kip-in.

LRFD

  0.75

  2.00

M n  0.75 1,500 kip-in.  1,130 kip-in.  731 kip-in.

o.k.

ASD

M n 1,500 kip-in.  2.00   750 kip-in.  488 kip-in.

o.k.

Shear Rupture Strength of Plate From AISC Specification Section J4.2(b), the available shear rupture strength of the plate is determined as follows: Anv  l  n  d h  z in.  t p  142 in.  5 18 in.  z in.   w in.  6.42 in.2 Rnv  0.60 Fu Anv



 0.60  65 ksi  6.42 in.2

(Spec. Eq. J4-4)



 250 kips

  0.75

LRFD

  2.00

Rnv  0.75  250 kips   188 kips  75 kips

ASD

Rnv 250 kips   2.00  125 kips  50 kips o.k.

o.k.

Interaction of Axial, Flexure and Shear Rupture in Plate AISC Specification Chapter H does not address combined flexure and shear. The method employed here is derived from Chapter H in conjunction with AISC Manual Equation 10-5, as follows: LRFD Nu 60 kips  Rnp 313 kips  0.192

ASD N a 40 kips  Rnp 209 kips  0.191

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LRFD

ASD N a  0.2 : Because Rnp

N Because u  0.2 : Rnp

 Nu Va  u   2Rnp M n

2

2

2

 N a Va a  Va  1    M n  Rnv  2 Rnp

  Vu      1   Rnv  2

 60 kips  75 kips  9w in.   75 kips       1 1,130 kip-in.   188 kips   2  313 kips  0.711  1 o.k. 2

2

 40 kips  50 kips  9w in.   50 kips       1 750 kip-in.   125 kips   2  209 kips  0.716  1 o.k. 2

Block Shear Rupture Strength of Plate—Beam Shear Direction The nominal strength for the limit state of block shear rupture of the plate, assuming an L-shaped tearout due to the shear load only as shown in Figure II.A-19B-2(a), is determined as follows:

Rbsv  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Agv   l  lev  t  142 in.  14 in. w in.  9.94 in.2 Anv  Agv   nv  0.5  d h  z in. t  9.94 in.2   5  0.5 18 in.  z in. w in.  5.93 in.2 Ant  leh   nh  1 s   nh  0.5  d h  z in.  t  1w in.   2  1 3 in.   2  0.518 in.  z in.   w in.  2.23 in.2

(a) Beam shear direction

(b) Beam axial direction— L-shaped Fig. II.A-19B-2. Block shear rupture of plate.

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(c) Beam axial direction— U-shaped

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Because stress is not uniform along the net tensile area, Ubs = 0.5.















Rbsv  0.60  65 ksi  5.93 in.2  0.5  65 ksi  2.23 in.2  0.60  50 ksi  9.94 in.2  0.5  65 ksi  2.23 in.2



 304 kips  371 kips

Therefore: Rbsv  304 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture on the plate is: LRFD

  0.75

Rbsv  0.75  304 kips   228 kips  75 kips

ASD

  2.00

Rbsv 304 kips   2.00  152 kips  50 kips

o.k .

o.k.

Block Shear Rupture Strength of the Plate—Beam Axial Direction The plate block shear rupture failure path due to axial load only could occur as an L- or U-shape. Assuming an Lshaped failure path due to axial load only, as shown in Figure II.A-19B-2(b), the available block shear rupture strength of the plate is:

Rbsn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Agv   nh  1 s  leh  t   2  1 3 in.  1w in.  w in.  3.56 in.2

Anv  Agv   nh  0.5  d h  z in. t  3.56 in.2   2  0.5 18 in.  z in  w in.  2.22 in.2 Ant  lev   nv  1 s   nv  0.5  d h  z in.  t  14 in.   5  1 3 in.   5  0.5 18 in.  z in    w in.  5.93 in.2 U bs  1.0

and















Rbsn  0.60  65 ksi  2.22 in.2  1.0  65 ksi  5.93 in.2  0.60  50 ksi  3.56 in.2  1.0  65 ksi  5.93 in.2  472 kips  492 kips

Therefore:

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Rbsn  472 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture on the plate is: LRFD

  0.75

Rbsn  0.75  472 kips   354 kips  60 kips

ASD

  2.00

Rbsn 472 kips   2.00  236 kips  40 kips

o.k .

o.k.

Assuming a U-shaped failure path in the plate due to axial load, as shown in Figure II.A-19B-2(c), the available block shear rupture strength of the plate is:

Rbsn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Agv   2 shear planes  leh   nh  1 s  t   2 shear planes  1w in.   2  1 3 in.   w in.  7.13 in.2

Anv  Agv   2 shear planes  nh  0.5  d h  z in. t  7.13 in.2   2 shear planes  2  0.518 in.  z in. w in.  4.46 in.2 Ant   nv  1 s   nv  1 d h  z in.  t   5  1 3 in.   5  118 in.  z in.   w in.  5.44 in.2 U bs  1.0

and















Rbsn  0.60  65 ksi  4.46 in.2  1.0  65 ksi  5.44 in.2  0.60  50 ksi  7.13 in.2  1.0  65 ksi  5.44 in.2



 528 kips  568 kips

Therefore: Rbsn  528 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture on the plate is:   0.75

LRFD

Rbsn  0.75  528 kips   396 kips  60 kips

  2.00

o.k .

ASD

Rbsn 528 kips   2.00  264 kips  40 kips

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Block Shear Rupture Strength of Plate—Combined Shear and Axial Interaction The same L-shaped block shear rupture failure path is loaded by forces in both the shear and axial directions. The interaction of loading in both directions is determined as follows: LRFD 2

ASD 2

2

 Va   N a      1  Rbsv   Rbsn 

 Vu   Nu      1  Rbsv   Rbsn  2

2

2

2

 75 kips   60 kips       0.137  1  228 kips   354 kips 

o.k.

2

 50 kips   40 kips       0.137  1  152 kips   236 kips 

o.k.

Shear Rupture Strength of Column Web at Weld From AISC Specification Section J4.2(b), the available shear rupture strength of the column web is determined as follows:

Anv  2ltw  2 142 in. 0.440 in.  12.8 in.2 Rn  0.60 Fu Av



 0.60  65 ksi  12.8 in.

2

(Spec. Eq. J4-4)



 499 kips

  0.75

LRFD

  2.00

Rn  0.75  499 kips   374 kips  75 kips

Rn 499 kips  2.00   250 kips  50 kips

o.k.

ASD

o.k.

Yield Line Analysis on Supporting Column Web A yield line analysis is used to determine the strength of the column web in the direction of the axial tension load. The yield line and associated dimensions are shown in Figure II.A-19B-3 and the available strength is determined as follows: T  d  2kdes  14.0 in.  2 1.31 in.  11.4 in.

d t  kdes  w 2 2 14.0 in. 0.415 in.   1.31 in.  2 2  5.90 in.

a

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d t  kdes  w  t p 2 2 14.0 in. 0.415 in.   1.31 in.   w in. 2 2  4.73 in.

b

c  tp  w in. tw2 Fy  4 2Tab  a  b   l  a  b     (Manual Eq. 9-31) 4  ab    2  0.440 in.  50 ksi   4 2 11.4 in. 5.90 in. 4.73 in. 5.90 in.  4.73 in.  142 in. 5.90 in.  4.73 in.   4  5.90 in. 4.73 in.      41.9 kips

Rn 

  1.00

LRFD

Rn  1.00  41.9 kips   41.9 kips  60 kips

n.g.

  1.50

ASD

Rn 41.9 kips   1.50  27.9 kips  40 kips

n.g.

The available column web strength is not adequate to resist the axial force in the beam. The column may be increased in size for an adequate web thickness or reinforced with stiffeners or web doubler plates. For example, a W14120 column, with tw = 0.590 in., has adequate strength to resist the given forces.

Fig II.A-19B-3. Yield line for column web.

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Strength of Weld A two-sided fillet weld with size of (s)tp = 0.469 in. (use 2-in. fillet welds) is used. As discussed in AISC Manual Part 10, this weld size will develop the strength of the shear plate used because the moment generated by this connection is indeterminate. The available weld strength is determined using AISC Manual Equation 8-2a or 8-2b, incorporating the directional strength increase from AISC Specification Equation J2-5, as follows:   1.0  0.50sin1.5   1.0  0.50sin1.5  38.7   1.25 LRFD Rn  1.392 kip/in. Dl  (2 sides)  1.392 kip/in. 8 142 in.1.25  2 sides   404 kips > 96.0 kips

o.k.

ASD Rn   0.928 kip/in. Dl  2 sides 

  0.928 kip/in. 8 142 in.1.25  2 sides   269 kips > 64.0 kips

o.k.

Conclusion The configuration given does not work due to the inadequate column web. The column would need to be increased in size or reinforced as discussed previously. Comments: If the applied axial load were in compression, the connection plate would need to be checked for compressive flexural buckling strength as follows. This is required in the case of the extended configuration of a single-plate connection and would not be required for the conventional configuration. From AISC Specification Table C-A-7.1, Case c: K  1.2 Lc KL  r r 1.2  9w in.  w in. 12  54.0

As stated in AISC Specification Section J4.4, if Lc/r is greater than 25, Chapter E applies. The available critical stress of the plate, Fcr or Fcr/, is determined using AISC Manual Table 4-14 as follows: LRFD

ASD

Fcr  36.4 ksi

Fcr  24.2 ksi 

Rn  Fcr lt p

Rn Fcr  lt p     24.2 ksi 142 in. w in.

  36.4 ksi 142 in. w in.  396 kips  60 kips

o.k.

 263 kips  40 kips

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Column Reinforcement As mentioned there are three options to correct the column web failure. These options are as follows: 1) Use a heavier column. This may not be practical because the steel may have been purchased and perhaps detailed and fabricated before the problem is found. 2) Use a web doubler plate. This plate would be fitted about the shear plate on the same side of the column web as the shear plate. This necessitates a lot of cutting, fitting and welding, and is therefore expensive. 3) Use stiffener or stabilizer plates—also called continuity plates. This is probably the most viable option, but changes the nature of the connection, because the stiffener plates will cause the column to be subjected to a moment. This cannot be avoided, but may be used advantageously. Option 3 Solution Because the added stiffeners cause the column to pick-up moment, the moment for which the connection is designed can be reduced. The connection is designed as a conventional configuration shear plate with axial force for everything to the right of Section A-A as shown in Figure II.A-19B-4. The design to the left of Section A-A is performed following a procedure for Type II stabilizer plates presented in Fortney and Thornton (2016). As shown in Figure II.A-19B-5, the moment in the shear plate to the left of Section A-A is uncoupled between the stabilizer plates.

Vs 

Va l

where a   7 in. l  142 in. g  2w in.

V a Vus  u l  75 kips  7 in.  142 in.  36.2 kips

LRFD

ASD V a Vas  a L 50  kips  7 in.  142 in.  24.1 kips

The force between the shear plate and stabilizer plate is determined as follows: LRFD N Fup  Vus  u 2  36.2 kips   66.2 kips

ASD Fap

60 kips 2

N  Vas  a 2  24.1 kips   44.1 kips

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40 kips 2

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Fig. II.A-19B-4. Design of shear plate with stabilizer plates.

Fig. II.A-19B-5. Forces acting on shear plate.

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Stabilizer Plate Design The stabilizer plate design is shown in Figure II.A-19B-6. The forces in the stabilizer plate are calculated as follows: LRFD Shear:

ASD Shear:

Fup 2 66.2 kips  2  33.1 kips

Vu 

Va 

Fap

2 44.1 kips  2  22.1 kips

Moment: Fup w Mu  4  66.2 kips 12 2in.  4  207 kip-in.

Moment: Fap w Ma  4  44.1 kips 12 2in.  4  138 kip-in.

Try s-in.-thick stabilizer plates. The available shear strength of the stabilizer plate is determined using AISC Specification Section J4.2 as follows:

Anv  bt   5w in. s in.  3.59 in.2 Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  65 ksi  3.59 in.

2



 140 kips

Fig. II.A-19B-6. Stabilizer plate design.

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  0.75

LRFD

  2.00

Rn  0.75 140 kips 

ASD

Rn 140 kips   2.00  70.0 kips  22.1 kips

 105 kips  33.1 kips o.k.

o.k.

The available flexural strength of the stabilizer plate is determined as follows: M n  Fy Z x   s in. 5w in.2     50 ksi   4    258 kip-in.

  0.90

LRFD

  1.67

M n  0.90  258 kip-in.  232 kip-in.  207 kip-in.

o.k.

ASD

M n 258 kip-in.   1.67  154 kip-in.  138 kip-in. o.k.

Stabilizer Plate to Column Weld Design The required weld size between the stabilizer plate and column flanges is determined using AISC Manual Equations 8-2a or 8-2b as follows:

Dreq

LRFD Fup 2   2 welds 1.392 kip/in. b 

Dreq

 66.2 kips 2   2 welds 1.392 kip/in. 5w in.

ASD Fap 2   2 welds  0.928 kip/in. b 

 2.07 sixteenths

 44.1 kips 2   2 welds  0.928 kip/in. 5w in.

 2.07 sixteenths

The minimum weld size per AISC Specification Table J2.4 controls. Use 4-in. fillet welds. Shear Plate to Stabilizer Plate Weld Design The required weld size between the shear plate and stabilizer plates is determined using AISC Manual Equations 82a or 8-2b as follows:

Dreq

LRFD Fup   2 welds 1.392 kip/in. lw 

Dreq

66.2 kips  2 welds 1.392 kip/in. 5w in.

 4.14 sixteenths

ASD Fap   2 welds  0.928 kip/in. lw 

44.1 kips  2 welds  0.928 kip/in. 5w in.

 4.13 sixteenths

Use c-in. fillet welds.

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Strength of Shear Plate at Stabilizer Plate Welds The minimum shear plate thickness that will match the shear rupture strength of the weld is: tmin  

6.19 D Fu

(Manual Eq. 9-3)

6.19  4.14 

65 ksi  0.394 in.  2 in. o.k.

Shear Plate to Column Web Weld Design The shear plate to stabilizer plate welds act as “crack arrestors” for the shear plate to column web welds. As shown in Figure II.A-19B-7, the required shear force is V. The required weld size is determined using AISC Manual Equations 8-2a or 8-2b as follows: LRFD

ASD

Vu  75 kips

Dreq  

Va  50 kips

Vu  2 welds 1.392 kip/in. l 75 kips

Dreq 

 2 welds 1.392 kip/in.142 in.

 1.86 sixteenths



Va  2 welds  0.928 kip/in. l 50 kips

 2 welds  0.928 kip/in.142 in.

 1.86 sixteenths

The minimum weld size per AISC Specification Table J2.4 controls. Use x-in. fillet welds.

Fig. II.A-19B-7. Moment induced in column.

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Strength of Shear Plate at Column Web Welds From AISC Specification Section J4.2(b), the available shear rupture strength of the shear plate is determined as follows:

Anv  lt  142 in.2 in.  7.25 in.2 Rn  0.60 Fu Anv



 0.60  65 ksi  7.25 in.

2

(Spec. Eq. J4-4)



 283 kips

  0.75

LRFD

  2.00

Rn  0.75  283 kips   212 kips  75 kips

Rn 283 kips   2.00  142 kips  50 kips

o.k.

ASD

o.k.

Moment in Column The moment in the column is determined as follows: LRFD 2 M u  Vus l

ASD 2 M a  Vas l

  36.2 kips 142 in.

  24.1 kips 142 in.

 525 kip-in.

 349 kip-in.

M u  263 kip-in.

M a  175 kip-in.

The column design needs to be reviewed to ensure that this moment does not overload the column.

Reference Fortney, P. and Thornton, W. (2016), “Analysis and Design of Stabilizer Plates in Single-Plate Shear Connections,” Engineering Journal, AISC, Vol. 53, No. 1, pp. 1–18.

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EXAMPLE II.A-20 ALL-BOLTED SINGLE-PLATE SHEAR SPLICE Given: Verify an all-bolted single-plate shear splice between two ASTM A992 beams, as shown in Figure II.A-20-1, to support the following beam end reactions: RD = 10 kips RL = 30 kips Use ASTM A36 plate.

Fig. II.A-20-1. Connection geometry for Example II.A-20.

Solution: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W2455 tw = 0.395 in.

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Beam W2468 tw = 0.415 in. From AISC Specification Table J3.3, for d-in.-diameter bolts with standard holes: dh = , in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2 10 kips   1.6  30 kips 

ASD Ra  10 kips  30 kips  40.0 kips

 60.0 kips Strength of the Bolted Connection—Plate

Note: When the splice is symmetrical, the eccentricity of the shear to the center of gravity of either bolt group is equal to half the distance between the centroids of the bolt groups. Therefore, each bolt group can be designed for the shear, Ru or Ra, and one-half the eccentric moment, Rue or Rae. Using a symmetrical splice, each bolt group will carry one-half the eccentric moment. Thus, the eccentricity on each bolt group is determined as follows: e 5 in.  2 2  2.50 in. From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the invividual strengths of the individual fasteners, which may be taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10 or the tearout strength at the bolt hole per AISC Specification Section J3.10. From AISC Manual Table 7-1, the available shear strength per bolt for d-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is: LRFD

ASD

rn = 16.2 kips/bolt 

rn = 24.3 kips/bolt

The available bearing strength of the plate per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration: (Spec. Eq. J3-6a)

rn  2.4dtFu   2.4  d in. a in. 58 ksi   45.7 kips/bolt

  0.75

LRFD

rn  0.75  45.7 kips/bolt   34.3 kips/bolt

  2.00 rn 45.7 kips/bolt  2.00   22.9 kips/bolt

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The available tearout strength of the plate per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration. Note: The available tearout strength based on edge distance will conservatively be used for all of the bolts.

lc  lev  0.5  d h   12 in.  0.5 , in.  1.03 in. rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2 1.03 in. a in. 58 ksi   26.9 kips/bolt

 = 0.75

LRFD

  2.00

rn = 0.75  26.9 kips/bolt 

ASD

rn 26.9 kips/bolt   2.00  13.5 kips/bolt

 20.2 kips/bolt

The tearout strength controls over bearing and shear for bolts in the plate. The available strength of the bolt group is determined by interpolating AISC Manual Table 7-6, with n = 4, Angle = 0, and ex = 22 in. C  3.07

Cmin

LRFD Ru  rn 60.0 kips  20.2 kips/bolt  2.97  3.07 o.k.

ASD Cmin

Ra  rn /  40.0 kips  13.5 kips/bolt  2.96  3.07 o.k.

Strength of the Bolted Connection—Beam Web By inspection, bearing and tearout on the webs of the beams will not govern. Flexural Yielding of Plate The required flexural strength is determined as follows: LRFD

ASD

Re Mu  u 2 60.0 kips  5 in.   2  150 kip-in.

Re Ma  a 2 40.0 kips  5 in.   2  100 kip-in.

The available flexural strength is determined as follows: LRFD

ASD

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 = 0.90

  1.67

M n  Fy Z x

M n Fy Z x   

  a in.12 in.2   0.90  36 ksi    4    437 kip-in.  150 kip-in. o.k.



2 36 ksi   a in.12 in.    1.67  4 

 291 kip-in.  100 kip-in. o.k.

Flexural Rupture of Plate The net plastic section modulus of the plate, Znet, is determined from AISC Manual Table 15-3:

Z net = 9.00 in.3 M n  Fu Z net

(Manual Eq. 9-4)



  58 ksi  9.00 in.

3



 522 kip-in.

LRFD

 = 0.75

  2.00

M n  0.75  522 kip-in.  392 kip-in.  150 kip-in.

ASD

522 kip-in. 2.00  261 kip-in.  100 kip-in.

M n 

o.k.

o.k.

Shear Strength of Plate From AISC Specification Section J4.2(a), the available shear yielding strength of the plate is determined as follows: Agv  lt  12 in. a in.  4.50 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  4.50 in.2



 97.2 kips

  1.00

LRFD

Rn  1.00  97.2 kips   97.2 kips  60.0 kips o.k.

  1.50

ASD

Rn 97.2 kips   1.50  64.8 kips  40.0 kips o.k.

From AISC Specification Section J4.2(b), the available shear rupture strength of the plate is determined using the net area determined in accordance with AISC Specification Section B4.3b.

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Anv   d  n  d h  z in.  t  12 in.  4 , in.  z in.   a in.  3.00 in.2

Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  58 ksi  3.00 in.2



 104 kips

  0.75

LRFD

ASD

  2.00

Rn  0.75 104 kips 

Rn 104 kips   2.00  52.0 kips  40.0 kips o.k.

 78.0 kips  60.0 kips o.k.

Block Shear Rupture of Plate The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the plate is determined as follows, using AISC Manual Tables 9-3a, 93b and 9-3c, and AISC Specification Equation J4-5, with n = 4, leh = lev = 12 in., and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a:  Fu Ant  43.5 kip/in.  t Shear yielding component from AISC Manual Table 9-3b:  0.60 Fy Agv  170 kip/in.   t

 Shear rupture component from AISC Manual Table 9-3c:



0.60Fu Anv  183 kip/in.  t

ASD Tension rupture component from AISC Manual Table 9-3a:





Fu Ant  29.0 kip/in. t

Shear yielding component from AISC Manual Table 9-3b:



0.60Fy Agv t

 113 kip/in.

Shear rupture component from AISC Manual Table 9-3c:



0.60Fu Anv  122 kip/in.  t

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LRFD Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant 

   a in. 183 kip/in.  1.0  43.5 kip/in.    a in. 170 kip/in.  1.0  43.5 kip/in. 

 84.9 kips  80.1 kips    Therefore: Rn  80.1 kips  60.0 kips

ASD Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant  +     a in. 122 kip/in.  1.0  29.0 kip/in. 

  a in. 113 kip/in.  1.0  29.0 kip/in.   56.6 kips  53.3 kips Therefore: o.k.

Rn  53.3 kips  40.0 kips o.k. 

Conclusion The connection is found to be adequate as given for the applied force.

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EXAMPLE II.A-21 BOLTED/WELDED SINGLE-PLATE SHEAR SPLICE Given:

Verify a single-plate shear splice between between two ASTM A992 beams, as shown in Figure II.A-21-1, to support the following beam end reactions: RD = 8 kips RL = 24 kips Use an ASTM A36 plate and 70-ksi electrodes.

Fig. II.A-21-1. Connection geometry for Example II.A-21. Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1631 tw = 0.275 in. Beam W1650 tw = 0.380 in.

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From AISC Specification Table J3.3, for w-in.-diameter bolts with standard holes: dh = m in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  8 kips   1.6  24 kips 

ASD Ra  8 kips  24 kips  32.0 kips

 48.0 kips

Strength of the Welded Connection—Plate Because the splice is unsymmetrical and the weld group is more rigid, it will be designed for the full moment from the eccentric shear. Use a PLa in.8 in.1 ft 0 in. This plate size meets the dimensional and other limitations of a single-plate connection with a conventional configuration from AISC Manual Part 10. Use AISC Manual Table 8-8 to determine the weld size. kl l 32 in.  12 in.  0.292

k

Interpolating from AISC Manual Table 8-8, with Angle = 0, and k = 0.292: x = 0.0538

xl   0.053812 in.  0.646 in. ex  al  6.50 in.  0.646 in.  5.85 in. al l 5.85 in.  12 in.  0.488

a

By interpolating AISC Manual Table 8-8, with Angle = 0: C = 2.15 From AISC Manual Equation 8-21, with C1 = 1.00 from AISC Manual Table 8-3, the required weld size is determined as follows:

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LRFD Dmin

ASD

Ru  CC1l 

Dmin

48.0 kips 0.75  2.15 1.00 12 in.

Ra  CC1l 

 2.48  3 sixteenths

 2.00  32.0 kips  2.15 1.00 12 in.

 2.48  3 sixteenths

The minimum required weld size from AISC Specification Table J2.4 is x in. Use a x-in. fillet weld. Shear Rupture of W1631 Beam Web at Weld For fillet welds with FEXX = 70 ksi on one side of the connection, the minimum thickness required to match the available shear rupture strength of the connection element to the available shear rupture strength of the base metal is: tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  2.48 

65 ksi  0.118 in.  0.275 in. o.k.

Strength of the Bolted Connection—Plate From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the invividual strengths of the individual fasteners, which may be taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is: LRFD

ASD

rn  11.9 kips/bolt 

rn  17.9 kips/bolt

The available bearing strength of the plate per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration: (Spec. Eq. J3-6a)

rn  2.4dtFu   2.4  w in. a in. 58 ksi   39.2 kips/bolt

  0.75

LRFD

rn  0.75  39.2 kips/bolt   29.4 kips/bolt

  2.00 rn 39.2 kips/bolt   2.00  19.6 kips/bolt

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The available tearout strength of the plate per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration. Note: The available tearout strength based on edge distance will conservatively be used for all of the bolts.

lc  lev  0.5  d h   12 in.  0.5 m in.  1.09 in. rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2 1.09 in. a in. 58 ksi   28.4 kips/bolt

 = 0.75

LRFD

  2.00

rn = 0.75  28.4 kips/bolt 

ASD

rn 28.4 kips/bolt   2.00  14.2 kips/bolt

 21.3 kips/bolt The bolt shear strength controls for bolts in the plate.

Because the weld group is designed for the full eccentric moment, the bolt group is designed for shear only.

nmin

LRFD Ru  rn 48.0 kips  17.9 kips/bolt  2.68 bolts  4 bolts o.k.

nmin

ASD Ra  rn /  32.0 kips  11.9 kips/bolt  2.69 bolts  4 bolts o.k.

Strength of the Bolted Connection—Beam Web By inspection, bearing and tearout on the W1650 beam web will not govern. Flexural Yielding of Plate The required flexural strength of the plate is determined as follows: LRFD

M u  Ru ex

ASD

M a  Ra ex

  48.0 kips  5.85 in.

  32.0 kips  5.85 in.

 281 kip-in.

 187 kip-in.

The available flexural strength of the plate is determined as follows:

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LRFD

 = 0.90

  1.67

M n  Fy Z x   a in.12 in.2   0.90  36 ksi    4    437 kip-in.  281 kip-in. o.k.

ASD

M n Fy Z x    

2 36 ksi   a in.12 in.    1.67  4 

 291 kip-in.  187 kip-in. o.k.

Shear Strength of Plate From AISC Specification Section J4.2(a), the available shear yielding strength of the plate is determined as follows: Agv  lt  12 in. a in.  4.50 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  4.50 in.

2



 97.2 kips

LRFD

  1.00

  1.50

Rn  1.00  97.2 kips 

ASD

Rn 97.2 kips   1.50  64.8 kips  32.0 kips o.k.

 97.2 kips  48.0 kips o.k.

From AISC Specification Section J4.2, the available shear rupture strength of the plate is determined using the net area determined in accordance with AISC Specification Section B4.3b.

Anv   d  n  d h  z in.  t  12 in.  4 m in.  z in.   a in.  3.19 in.2

Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  58 ksi  3.19 in.2



 111 kips

  0.75

LRFD

Rn  0.75 111 kips   83.3 kips  48.0 kips o.k.

  2.00

ASD

Rn 111 kips   2.00  55.5 kips  32.0 kips o.k.

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Block Shear Rupture of Plate The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the plate is determined as follows, using AISC Manual Tables 9-3a, 9-3b and 9-3c, and AISC Specification Equation J4-5, with n = 4, leh = lev = 12 in., and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a:  Fu Ant  46.2 kip/in.  t Shear yielding component from AISC Manual Table 9-3b:  0.60 Fy Agv  170 kip/in.   t

ASD Tension rupture component from AISC Manual Table 9-3a:





Fu Ant  30.8 kip/in. t

Shear yielding component from AISC Manual Table 9-3b:



0.60 Fy Agv t

 113 kip/in. 





Shear rupture component from AISC Manual Table 9-3c:

Shear rupture component from AISC Manual Table 9-3c:



0.60 Fu Anv  194 kip/in.  t

 



Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant   a in. 194 kip/in.  1.0  46.2 kip/in.    a in. 170 kip/in.  1.0  46.2 kip/in.   90.1 kips  81.1 kips





Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant  +     a in. 129 kip/in.  1.0  30.8 kip/in.    a in. 113 kip/in.  1.0  30.8 kip/in.   59.9 kips  53.9 kips

Therefore: Rn  81.1 kips  48.0 kips

0.60Fu Anv  129 kip/in. t

Therefore: o.k.

Rn  53.9 kips  32.0 kips o.k.  

Conclusion The connection is found to be adequate as given for the applied force.

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EXAMPLE II.A-22 BOLTED BRACKET PLATE DESIGN Given:

Verify the bracket plate to support the loads as shown in Figure II.A-22-1 (loads are per bracket plate). Use ASTM A36 plate. Assume the column has sufficient available strength for the connection.

Fig. II.A-22-1. Connection geometry for Example II.A-22. Solution:

For discussion of the design of a bracket plate, see AISC Manual Part 15. From AISC Manual Table 2-5, the material properties are as follows: Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2  6 kips   1.6 18 kips   36.0 kips

ASD Pa  6 kips  18 kips  24.0 kips

From the geometry shown in Figure II.A-22-1 and AISC Manual Figure 15-2(b):

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a  20 in. b  154 in. e  94 in. b   tan 1   a  154 in.   tan 1    20 in.   37.3 a cos  20 in.  cos 37.3  25.1 in.

a 

(Manual Eq. 15-17)

b  a sin    20 in. sin 37.3   12.1 in. Strength of the Bolted Connection—Plate From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the invividual strengths of the individual fasteners, which may be taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is: LRFD

ASD

rn  11.9 kips/bolt 

rn  17.9 kips/bolt

The available bearing and tearout strength of the plate is determined using AISC Manual Table 7-5 conservatively using le = 2 in. Note: The available bearing and tearout strength based on edge distance will conservatively be used for all of the bolts. LRFD

ASD rn   52.2 kip/in. a in.   19.6 kips/bolt

rn   78.3 kip/in. a in.  29.4 kips/bolt Bolt shear strength controls for bolts in the plate.

The strength of the bolt group is determined by interpolating AISC Manual Table 7-8 with Angle = 00, a 52 in. gage with s = 3 in., ex = 12 in. and n = 6:

C = 4.53

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Cmin

LRFD Pu  rn 36.0 kips  17.9 kips/bolt  2.01  4.53 o.k.

ASD Cmin

Pa  rn 24.0 kips  11.9 kips/bolt  2.02  4.53 o.k.

Flexural Yielding of Bracket Plate on Section K-K The required flexural yielding strength of the plate at Section K-K is determined from AISC Manual Equation 15-1a or 15-1b as follows: LRFD

ASD

M u  Pu e

M a  Pa e

  36.0 kips  94 in.

  24.0 kips  94 in.

 333 kip-in.

 222 kip-in.

The available flexural yielding strength of the bracket plate is determined as follows: M n  Fy Z

(Manual Eq. 15-2)

  a in. 20 in.2     36 ksi   4    1,350 kip-in.

LRFD

  0.90

  1.67

M n  0.90 1,350 kip-in.  1, 220 kip-in.  333 kip-in. o.k.

ASD

M n 1,350 kip-in.   1.67  808 kip-in.  222 kip-in.

o.k.

Flexural Rupture of Bracket Plate on Section K-K From AISC Manual Table 15-3, for a a-in.-thick bracket plate, with w-in. bolts and six bolts in a row, Znet = 21.5 in.3 Note that AISC Manual Table 15-3 conservatively considers lev  12 in. for holes spaced at 3 in. The available flexural yielding rupture of the bracket plate at Section K-K is determined as follows: M n  Fu Z net



  58 ksi  21.5 in.3

(Manual Eq. 15-3)



 1, 250 kip-in.

  0.75

LRFD

  2.00

M n  0.75 1, 250 kip-in.  938 kip-in.  333 kip-in. o.k.

ASD

M n 1, 250 kip-in.   2.00  625 kip-in.  222 kip-in.

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Shear Yielding of Bracket Plate on Section J-J The required shear strength of the bracket plate on Section J-J is determined from AISC Manual Equation 15-6a or 15-6b as follows: LRFD Vu  Pu sin 

ASD Va  Pa sin 

  36.0 kips  sin 37.3 

  24.0 kips  sin 37.3 

 21.8 kips

 14.5 kips

The available shear yielding strength of the plate is determined as follows:

Vn  0.6 Fy tb

(Manual Eq. 15-7)

 0.6  36 ksi  a in.12.1 in.  98.0 kips LRFD

  1.00

Vn  1.00  98.0 kips   98.0 kips  21.8 kips

o.k.

  1.50

ASD

Vn 98.0 kips   1.50  65.3 kips  14.5 kips o.k.

Local Yielding and Local Buckling of Bracket Plate on Section J-J (see Figure II.A-22-1) For local yielding:

Fcr  Fy

(Manual Eq. 15-13)

 36 ksi For local buckling:

Fcr  QFy

(Manual Eq. 15-14)

where





 b    Fy t   b  5 475  1,120    a 

(Manual Eq. 15-18)

2

 12.1 in.    36 ksi  a in. 

 12.1 in.  5 475  1,120    25.1 in.   1.43

2

Because 1.41< :

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Q

1.30



(Manual Eq. 15-16)

2 1.30

1.432

 0.636 Fcr  QFy

(Manual Eq. 15-14)

 0.636  36 ksi   22.9 ksi

Local buckling controls over local yielding. Interaction of Normal and Flexural Strengths Check that Manual Equation 15-10 is satisfied: LRFD N u  Pu cos 

ASD (Manual Eq. 15-9a)

N a  Pa cos 

(Manual Eq. 15-9b)

  36.0 kips  cos 37.3 

  24.0 kips  cos 37.3 

 28.6 kips

 19.1 kips

N n  Fcr tb    22.9 ksi  a in.12.1 in.

(Manual Eq. 15-11)

 104 kips

N n  Fcr tb    22.9 ksi  a in.12.1 in.

 104 kips

  0.90

  1.67

 N c  N n

Nc 

Nn  104 kips  1.67  62.3 kips

 0.90 104 kips    93.6 kips    b  M u  Pu e  N u   2

(Manual Eq. 15-8a)

 12.1 in.    36.0 kips  94 in.   28.6 kips     2   160 kip-in.

Mn  

Fcr tb2 4

(Manual Eq. 15-12)

 22.9 ksi  a in.12.1 in.2

 314 kip-in.

(Manual Eq. 15-11)

4

 b  M a  Pa e  N a   2

(Manual Eq. 15-8b)

 12.1 in.    24.0 kips  94 in.  19.1 kips     2   106 kip-in.

Mn  

Fcr tb2 4

(Manual Eq. 15-12)

 22.9 ksi  a in.12.1 in.2

 314 kip-in.

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LRFD

ASD

M c  M n

M Mc  n  314 kip-in.  1.67  188 kip-in.

 0.90  314 kip-in.  283 kip-in.

Nr M r (Manual Eq. 15-10)   1.0 Nc M c 28.6 kips 160 kip-in.   0.871  1.0 o.k. 93.6 kips 283 kip-in.

Nr M r (Manual Eq. 15-10)   1.0 Nc M c 19.1 kips 106 kip-in.   0.870  1.0 o.k. 62.3 kips 188 kip-in.

Shear Strength of Bracket Plate on Section K-K From AISC Specification Section J4.2, the available shear yielding strength of the plate on Section K-K is determined as follows: Agv  at   20 in. a in.  7.50 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  7.50 in.2



 162 kips

LRFD

  1.00

  1.50

Rn  1.00 162 kips 

ASD

Rn 162 kips   1.50  108 kips  24.0 kips o.k.

 162 kips  36.0 kips o.k.

From AISC Specification Section J4.2, the available shear rupture strength of the plate on Section K-K is determined as follows: Anv   a  n , in. + z in.  t   20 in.  6 m in. + z in.   a in.  5.53 in.2 Rn  0.60 Fu Anv



 0.60  58 ksi  5.53 in.

2

(Spec. Eq. J4-4)



 192 kips

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  0.75

LRFD

Rn  0.75 192 kips   144 kips  36.0 kips o.k.

  2.00

ASD

Rn 192 kips   2.00  96.0 kips  24.0 kips o.k.

Conclusion The connection is found to be adequate as given for the applied force.

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EXAMPLE II.A-23 WELDED BRACKET PLATE DESIGN Given:

Verify the welded bracket plate to support the loads as shown in Figure II.A-23-1 (loads are resisted equally by the two bracket plates). Use ASTM A36 plate and 70-ksi electrodes. Assume the column has sufficient available strength for the connection.

Fig. II.A-23-1. Connection geometry for Example II.A-23. Solution:

From AISC Manual Table 2-5, the material properties are as follows: Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From ASCE/SEI 7, Chapter 2, the required strength to be resisted by the bracket plates is: LRFD Pu  1.2  9 kips   1.6  27 kips   54.0 kips

ASD Pa  9 kips  27 kips  36.0 kips

From the geometry shown in Figure II.A-23-1 and AISC Manual Figure 15-2(b):

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a  18 in. b  112 in. e  84 in. b   tan 1   a  112 in.   tan 1    18 in.   32.6 a cos  18 in.  cos 32.6  21.4 in.

a 

(Manual Eq. 15-17)

b  a sin   18 in. sin 32.6   9.70 in. Shear Yielding of Bracket Plate at Section A-A From AISC Specification Section J4.2(a), the available shear yielding strength of the bracket plate at Section A-A, is determined as follows: Agv   2 plates  at   2 plates 18 in. a in.  13.5 in.2

Rn  0.60 Fy Agv



 0.60  36 ksi  13.5 in.2

(Spec. Eq. J4-3)



 292 kips

  1.00

LRFD

  1.50

Rn  1.00  292 kips 

ASD

Rn 292 kips   1.50  195 kips  36.0 kips o.k.

 292 kips  54.0 kips o.k. Shear rupture strength is adequate by insection. Flexural Yielding of Bracket Plate at Section A-A

The required flexural strength of the bracket plate is determined using AISC Manual Equation 15-1a or 15-1b as follows:

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LRFD

ASD

M u  Pu e

M a  Pa e

  54.0 kips  84 in.

  36.0 kips  84 in.

 446 kip-in.

 297 kip-in.

The available flexural strength of the bracket plate is determined using AISC Manual Equation 15-2, as follows: M n   2 plates  Fy Z

(from Manual Eq. 15-2)

  a in.18 in.   2 plates  36 ksi   4   2,190 kip-in.

  0.90

2

  

LRFD

  1.67

M n  0.90  2,190 kip-in.  1,970 kip-in.  446 kip-in.

o.k.

ASD

M n 2,190 kip-in.   1.67  1,310 kip-in.  297 kip-in.

Weld Strength Try a C-shaped weld with kl = 3 in. and l = 18 in.

kl l 3 in.  18 in.  0.167

k

 kl 2 2  kl   l  3 in.2  2  3 in.  18 in.

xl 

 0.375 in.

al  114 in.  0.375 in.  10.9 in. al l 10.9 in.  18 in.  0.606

a

Interpolate AISC Manual Table 8-8 using Angle = 00, k = 0.167, and a = 0.606. C  1.46

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From AISC Manual Table 8-3: C1  1.00 (for E70 electrodes)

The required weld size is determined using AISC Manual Equation 8-21, as follows:   0.75

Dmin  

LRFD

ASD

  2.00

Pu CC1l

Dmin 

54.0 kips 0.75 1.46 1.00 18 in. 2 plates 



Pa CC1l

2.00  36.0 kips 

1.46 1.00 18 in. 2 plates 

 1.37  3 sixteenths

 1.37  3 sixteenths

From AISC Specification Section J2.2b(b)(2), the maximum weld size is: wmax  a in.  z in.  c in.  x in.

o.k.

From AISC Specification Table J2.4, the minimum weld size is: wmin  x in.

Shear Yielding Strength of Bracket at Section B-B The required shear strength of the bracket plate at Section B-B is determined from AISC Manual Equations 15-6a or 15-6b as follows: LRFD

ASD

Vu  Pu sin 

Va  Pa sin 

  54.0 kips  sin 32.6 

  36.0 kips  sin 32.6 

 29.1 kips

 19.4 kips

From AISC Manual Part 15, the available shear yielding strength of the bracket plate at Section A-A is determined as follows:

Vn   2 plates  0.6Fy tb

(from Manual Eq. 15-7)

  2 plates  0.6  36 ksi  a in. 9.70 in.  157 kips   1.00

LRFD

  1.50

Vn  1.00 157 kips 

ASD

Vn 157 kips   1.50  105 kips  19.4 kips o.k.

 157 kips  29.1 kips o.k.

Bracket Plate Normal and Flexural Strength at Section B-B

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From AISC Manual Part 15, the required strength of the bracket plate at Section B-B is determined as follows: LRFD N u  Pu cos 

ASD (Manual Eq. 15-9a)

N a  Pa cos 

  54.0 kips  cos 32.6 

  36.0 kips  cos 32.6 

 45.5 kips

 30.3 kips

 b  M u  Pu e  N u   2

(Manual Eq. 15-8a)

 9.70 in.    54.0 kips  84 in.   45.5 kips     2   225 kip-in.

 b  M a  Pa e  N a   2

(Manual Eq. 15-9b)

(Manual Eq. 15-8b)

 9.70 in.    36.0 kips  84 in.   30.3 kips     2   150 kip-in.

For local yielding at the bracket plate:

Fcr  Fy

(Manual Eq. 15-13)

 36 ksi For local buckling of the bracket plate:

Fcr  QFy

(Manual Eq. 15-14)

where 



 b    Fy t 

(Manual Eq. 15-18)

2

 b  5 475  1,120    a   9.70 in.    36 ksi  a in. 

 9.70 in.  5 475  1,120    21.4 in.   1.17

2

Since 0.70    1.41: Q  1.34  0.486

(Manual Eq. 15-15)

 1.34  0.486 1.17   0.771 Fcr  QFy

(Manual Eq. 15-14)

 0.771 36 ksi   27.8 ksi

Therefore; local buckling governs over yielding. The nominal strength of the bracket plate for the limit states of local yielding and local buckling is:

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N n   2 plates  Fcr tb

(from Manual Eq. 15-11)

  2 plates  27.8 ksi  a in. 9.70 in.  202 kips

The nominal flexural strength of the bracket plate for the limit states of local yielding and local buckling is:

M n   2 plates    2 plates 

Fcr tb2 4

(from Manual Eq. 15-12)

 27.8 ksi  a in. 9.70 in.2 4

 490 kip-in. LRFD

ASD

Mr  Mu  225 kip-in.

Mr  Ma  150 kip-in.

  0.90

  1.67

M c  M n

Mc 

 0.90  490 kip-in.  441 kip-in.  225 kip-in.

o.k.

N r  Nu  45.5 kips

o.k.

Nr  Na  30.3 kips

N c  N n

Nn  202 kips  1.67  121 kips  30.3 kips

Nc 

 0.90  202 kips   182 kips  45.5 kips

Mn  490 kip-in.  1.67  293 kip-in.  150 kip-in.

o.k.

Nr M r   1.0 (Manual Eq. 15-10) Nc M c 45.5 kips 225 kip-in.   0.760  1.0 o.k. 182 kips 441 kip-in.

Nr M r   1.0 Nc M c

o.k.

(Manual Eq. 15-10)

30.3 kips 150 kip-in.   0.762  1.0 121 kips 293 kip-in.

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EXAMPLE II.A-24 ECCENTRICALLY LOADED BOLT GROUP (IC METHOD) Given:

Use AISC Manual Table 7-8 to determine the largest eccentric force, acting vertically (0 angle) and at a 15° angle, which can be supported by the available shear strength of the bolts using the instantaneous center of rotation method. Assume that bolt shear controls over bearing and tearout. Solution A (  0°):

Assume the load is vertical ( = 00), as shown in Figure II.A-24-1: From AISC Manual Table 7-1, the available shear strength per bolt for d-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in single shear is: LRFD

ASD

rn  16.2 kips/bolt 

rn  24.3 kips/bolt 

The available strength of the bolt group is determined using AISC Manual Table 7-8, with Angle = 00, a 52-in. gage with s = 3 in., ex = 16 in., and n = 6: C  3.55

LRFD Rn  C rn

ASD Rn rn C    3.55 16.2 kips/bolt 

(Manual Eq. 7-16)

 3.55  24.3 kips/bolt   86.3 kips

(Manual Eq. 7-16)

 57.5 kips



 Thus, Pu must be less than or equal to 86.3 kips.

Thus, Pa must be less than or equal to 57.5 kips.

Fig. II.A-24-1. Connection geometry for Example II.A-24—Solution A (  0).

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Note: The eccentricity of the load significantly reduces the shear strength of the bolt group. Solution B (  15°):

Assume the load acts at an angle of 150 with respect to vertical ( = 150), as shown in Figure II.A-24-2: ex  16 in.   9 in. tan15   18.4 in.

The available strength of the bolt group is determined interpolating from AISC Manual Table 7-8, with Angle = 150, a 52-in. gage with s = 3 in., ex = 18.4 in., and n = 6: C  3.21

LRFD Rn  C rn

(Manual Eq. 7-16)

 3.21 24.3 kips/bolt   78.0 kips

ASD Rn rn C    3.2116.2 kips/bolt 

(Manual Eq. 7-16)

 52.0 kips Thus, Pu must be less than or equal to 78.0 kips.

Thus, Pa must be less than or equal to 52.0 kips.

Fig. II.A-24-2. Connection geometry for Example II.A-24—Solution B (  15).

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EXAMPLE II.A-25 ECCENTRICALLY LOADED BOLT GROUP (ELASTIC METHOD) Given:

Determine the largest eccentric force that can be supported by the available shear strength of the bolts using the elastic method for  = 0, as shown in Figure II.A-25-1. Compare the result with that of Example II.A-24. Assume that bolt shear controls over bearing and tearout.

Fig. II.A-25-1. Connection geometry for Example II.A-25. Solution:

From AISC Manual Table 7-1, the available shear strength per bolt for d-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in single shear is: LRFD

ASD

rn  16.2 kips/bolt 

rn  24.3 kips/bolt 

The direct shear force per bolt is determined as follows: LRFD

ASD

rpxu  0

Pu n Pu  12

rpyu 

rpxa  0 (from Manual Eq. 7-2a)

Pa n Pa  12

rpya 

Additional shear force due to eccentricity is determined as follows: The polar moment of inertia of the bolt group is:

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I x  y 2  4  7.50 in. + 4  4.50 in. + 4 1.50 in. 2

2

2

 315 in.4 /in.2 I y  x 2  12  2.75 in.

2

 90.8 in.4 /in.2

I p  Ιx  I y  315 in.4 /in.2  90.8 in.4 /in.2  406 in.4 /in.2 LRFD

rmxu

ASD

Pu ec y  Ip 

(Manual Eq. 7-6a)

rmxa

Pu 16.0 in. 7.50 in. 4



2



Pu ecx Ip

(Manual Eq. 7-7a)

rmya 

Pu 16.0 in. 2.75 in. 4



2

The resultant shear force is determined from AISC Manual Equation 7-8a:



 rpxu  rmxu    rpyu  rmyu  2

 0  0.296 Pu 2  

2

Pu   0.108Pu   12 

2

(Manual Eq. 7-7b)

Pa 16.0 in. 2.75 in.



 rpxa  rmxa    rpya  rmya  2

 0  0.296 Pa 2  

2

Pa   0.108Pa   12 

2

 0.352 Pa

Because ru must be less than or equal to the available strength:

rn 0.352 24.3 kips/bolt  0.352  69.0 kips

Pa ecx Ip

The resultant shear force is determined from AISC Manual Equation 7-8b: ra 

 0.352 Pu

Pu 

Pa 16.0 in. 7.50 in.

406 in.4 /in.2  0.108Pa

406 in. /in.  0.108Pu

ru 

(Manual Eq. 7-6b)

406 in.4 /in.2  0.296 Pa

406 in. /in.  0.296 Pu rmyu 

Pa ec y  Ip

Because ra must be less than or equal to the available strength:

rn  0.352 16.2 kips/bolt  0.352  46.0 kips

Pa 

Note: The elastic method, shown here, is more conservative than the instantaneous center of rotation method, shown in Example II.A-24.

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EXAMPLE II.A-26 ECCENTRICALLY LOADED WELD GROUP (IC METHOD) Given:

Use AISC Manual Table 8-8 to determine the largest eccentric force, acting vertically and at a 75° angle, that can be supported by the available shear strength of the weld group, using the instantaneous center of rotation method. Use a a-in. fillet weld and 70-ksi electrodes. Solution A ( = 0°):

Assume that the load is vertical ( = 0°), as shown in Figure II.A-26-1.

kl l 5 in.  10 in.  0.500

k

 kl 2 2  kl   l  5 in.2  2  5 in.  10 in.

xl 

 1.25 in. xl  al  10.0 in. 1.25 in.  a 10 in.  10 in. a  0.875 ex  al  0.875 10 in.  8.75 in.

Fig. II.A-26-1. Weld geometry—Solution A (  0).

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The available weld strength is determined using AISC Manual Equation 8-21 and interpolating AISC Manual Table 8-8, with Angle = 0, a = 0.875, and k = 0.5: C  1.88 C1  1.00 (from AISC Manual Table 8-3)

Rn  CC1 Dl

(Manual Eq. 8-21)

 1.88 1.00  6 10 in.  113 kips

  0.75

LRFD

  2.00

Rn  0.75 113 kips 

ASD

Rn 113 kips  2.00   56.5 kips

 84.8 kips

Thus, Pu must be less than or equal to 84.8 kips.

Thus, Pa must be less than or equal to 56.5 kips.

Note: The eccentricity of the load significantly reduces the shear strength of this weld group as compared to the concentrically loaded case. Solution B ( = 75°):

Assume that the load acts at the same point as in Solution A, but at an angle of 75° with respect to vertical ( = 75°) as shown in Figure II.A-26-2. As determined in Solution A:

k  0.500 a  0.875

Fig. II.A-26-2. Weld geometry—Solution B (  75).

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The available weld strength is determined using AISC Manual Equation 8-21 and interpolating AISC Manual Table 8-8, with Angle = 75o, a = 0.875, and k = 0.5: C  3.45 C1  1.00 (from AISC Manual Table 8-3)

Rn  CC1 Dl

(Manual Eq. 8-21)

 3.45 1.00  6 10 in.  207 kips

  0.75

LRFD

  2.00

Rn  0.75  207 kips 

ASD

Rn 207 kips   2.00  104 kips

 155 kips Thus, Pu must be less than or equal to 155 kips.

Thus, Pa must be less than or equal to 104 kips.

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EXAMPLE II.A-27 ECCENTRICALLY LOADED WELD GROUP (ELASTIC METHOD) Given:

Using the elastic method determine the largest eccentric force that can be supported by the available shear strength of the welds in the connection shown in Figure II.A-27-1. Compare the result with that of Example II.A-26. Use ain. fillet welds and 70-ksi electrodes.

Fig. II.A-27-1. Weld geometry for Example II.A-27. Solution:

From the weld geometry shown in Figure II.A-27-1 and AISC Manual Table 8-8:

kl l 5 in.  10 in.  0.500

k

 kl 2 2  kl   l  5 in.2  2  5 in.  10 in.

xl 

 1.25 in. xl  al  10.0 in. 1.25 in.  a 10 in.  10 in. a  0.875 ex  al  0.875 10 in.  8.75 in.

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Direct Shear Force Per Inch of Weld LRFD

ASD

rpux  0

rpax  0

Pu

rpuy 

(from Manual Eq. 8-5a)

ltotal Pu  20.0 in. 0.0500 Pu  in.

rpay 

Pa

(from Manual Eq. 8-5b)

ltotal Pa  20.0 in. 0.0500 Pa  in.

Additional Shear Force due to Eccentricity Determine the polar moment of inertia referring to the AISC Manual Figure 8-6: Ix 

l3 2  2  kl  y  12



10 in.3

 2  5 in. 5 in.

2

12  333 in.4 /in.

2   kl 3 2  kl     kl    xl    l  xl  Iy  2 12 2       5 in.3  2 2  2   5 in. 2.50 in.  14 in.   10 in.14 in.  12 

 52.1 in.4 /in.

I p  Ix  I y  333 in.4 /in.  52.1 in.4 /in.  385 in.4 /in. LRFD rmux  

Pu ex c y Ip

ASD (from Manual Eq. 8-9a)

Pu  8.75 in. 5 in. 4

385 in. /in. 0.114 Pu  in.

rmax  

Pa ex c y Ip Pa  8.75 in. 5 in.

385 in.4 /in. 0.114 Pa  in.

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LRFD rmuy

Pe c  u x x Ip 

ASD

(from Manual Eq. 8-10a)

rmay

Pu  8.75 in. 3.75 in.



4

The resultant shear force is determined using AISC Manual Equation 8-11a:

 rpux  rmux    rpuy  rmuy  2

2

2

Pa  8.75 in. 3.75 in.

The resultant shear force is determined using Manual Equation 8-11b: ra 

0.114 Pu   0.0500 Pu 0.0852 Pu    0      in.   in. in.   0.177 Pu  in.

2

Because ru must be less than or equal to the available strength:

ru 

(from Manual Eq. 8-10b)

385 in.4 /in. 0.0852 Pa  in.

385 in. /in. 0.0852 Pu  in.

ru 

Pe c  a x x Ip

0.177 Pu  rn in.

 rpax  rmax    rpay  rmay  2

2

2

0.114 Pa   0.0500 Pa 0.0852 Pa    0      in.   in. in.   0.177 Pa  in.

2

Because ra must be less than or equal to the available strength:

ra 

0.177 Pa rn  in. 

Solving for Pu and using AISC Manual Equation 8-2a:

Solving for Pa and using AISC Manual Equation 8-2b:

 in.  Pu  rn    0.177 

Pa 

 in.   1.392 kip/in. 6     0.177   47.2 kips

rn  in.      0.177 

 in.    0.928 kip/in. 6     0.177   31.5 kips

Note: The strength of the weld group calculated using the elastic method, as shown here, is significantly less than that calculated using the instantaneous center of rotation method in Example II.A-26.

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EXAMPLE II.A-28A

ALL-BOLTED SINGLE-ANGLE CONNECTION (BEAM-TO-GIRDER WEB)

Given:

Verify an all-bolted single-angle connection (Case I in AISC Manual Table 10-11) between an ASTM A992 W1835 beam and an ASTM A992 W2162 girder web, as shown in Figure II.A-28A-1, to support the following beam end reactions: RD = 6.5 kips RL = 20 kips The top flange is coped 2 in. deep by 4 in. long, lev = 12 in., and leh = 1w in. Use ASTM A36 angle. Use standard angle gages.

Fig. II.A-28A-1. Connection geometry for Example II.A-28A. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beam and girder ASTM A992 Fy = 50 ksi Fu = 65 ksi Angle ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1 the geometric properties are as follows: Beam W1835

tw = 0.300 in. d = 17.7 in.

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tf = 0.425 in. Girder W2162

tw = 0.400 in. From AISC Specification Table J3.3, for w-in.-diameter bolts with standard holes: dh = m in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  6.5 kips   1.6  20 kips 

ASD Ra  6.5 kips  20 kips  26.5 kips

 39.8 kips

Strength of the Bolted Connection—Angle Check eccentricity of connection. For the 4-in. angle leg attached to the supported beam (W1835): e = 22 in. < 3.00 in., therefore, eccentricity does not need to be considered for this leg. (See AISC Manual Figure 10-14) For the 3-in. angle leg attached to the supporting girder (W2162): e  1w in. 

0.300 in. 2

 1.90 in. Because e = 1.90 in. < 22 in., AISC Manual Table 10-11 may be conservatively used for bolt shear. From Table 10-11, Case I, with n = 4: C  3.07 From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the invividual strengths of the individual fasteners, taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. In this case, the 3-in. angle leg attached to the supporting girder will control because eccentricity must be taken into consideration and the available strength will be determined based on the bolt group using the eccentrically loaded bolt coefficient, C. From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is: LRFD

rn  17.9 kips/bolt

ASD

rn  11.9 kips/bolt 

The available bearing and tearout strength of the angle at the bottom edge bolt is determined using AISC Manual Table 7-5, with le = 14 in., as follows:

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LRFD

ASD rn   29.4 kip/in. a in.   11.0 kips/bolt

rn   44.0 kip/in. a in.  16.5 kips/bolt

The available bearing and tearout strength of the angle at the interior bolts (not adjacent to the edge) is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

ASD rn   52.2 kip/in. a in.   19.6 kips/bolt

rn   78.3 kip/in. a in.  29.4 kips/bolt

The available strength of the bolted connection at the angle is conservatively determined using the minimum available strength calculated for bolt shear, bearing on the angle, and tearout on the angle. The bolt group eccentricity is accounted for by multiplying the minimum available strength by the bolt coefficient C. LRFD

ASD

Rn r C n    3.07 11.0 kips/bolt 

Rn  Crn  3.07 16.5 kips/bolt   50.7 kips  39.8 kips o.k.

 33.8 kips  26.5 kips o.k.

Strength of the Bolted Connection—W1835 Beam Web The available bearing and tearout strength of the beam web at the top edge bolt is determined using AISC Manual Table 7-5, conservatively using le = 14 in., as follows: LRFD

rn   49.4 kip/in. 0.300 in.  14.8 kips/bolt

ASD rn   32.9 kip/in. 0.300 in.   9.87 kips/bolt

The available bearing and tearout strength of the beam web at the interior bolts (not adjacent to the edge) is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

rn   87.8 kip/in. 0.300 in.  26.3 kips/bolt

ASD rn   58.5 kip/in. 0.300 in.   17.6 kips/bolt

The available strength of the bolted connection at the beam web is determined by summing the effective strength for each bolt using the minimum available strength calculated for bolt shear, bearing on the web, and tearout on the web.

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LRFD

ASD Rn rn n    1 bolt  9.87 kips/bolt 

Rn  nrn  1 bolt 14.8 kips/bolt    3 bolts 17.9 kips/bolt 

  3 bolts 11.9 kips/bolt 

 68.5 kips  39.8 kips o.k.

 45.6 kips  26.5 kips o.k.

Strength of the Bolted Connection—W2162 Girder Web The available bearing and tearout strength of the girder web is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

ASD rn   58.5 kip/in. 0.400 in.   23.4 kips/bolt

rn   87.8 kip/in. 0.400 in.  35.1 kips/bolt

Therefore; bolt shear controls over bearing or tearout on the girder web and is adequate based on previous calculations. Shear Strength of Angle From AISC Specification Section J4.2(a), the available shear yielding strength of the angle is determined as follows: Agv  lt  112 in. a in.  4.31 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  4.31 in.2



 93.1 kips

  1.00

LRFD

  1.50

Rn  1.00  93.1 kips 

ASD

Rn 93.1 kips   1.50  62.1 kips  26.5 kips o.k.

 93.1 kips  39.8 kips o.k.

From AISC Specification Section J4.2(b), the available shear rupture strength of the angle is determined using the net area determined in accordance with AISC Specification Section B4.3b.

Anv  l  n  d h  z in.  t  112 in.  4 m in.  z in.   a in.  3.00 in.2

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Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  58 ksi  3.00 in.

2



 104 kips

  0.75

LRFD

ASD

  2.00

Rn  0.75 104 kips 

Rn 104 kips   2.00  52.0 kips  26.5 kips o.k.

 78.0 kips  39.8 kips o.k. Block Shear Rupture of Angle

The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the 3-in. leg is determined as follows, using AISC Manual Tables 9-3a, 9-3b and 9-3c, and AISC Specification Equation J4-5, with n = 4, lev = leh = 14 in., and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a:  Fu Ant  35.3 kip/in.  t Shear yielding component from AISC Manual Table 9-3b:  0.6 Fy Agv  166 kip/in.   t

 Shear rupture component from AISC Manual Table 9-3c:



0.6 Fu Anv  188 kip/in.  t

  a in. 188 kip/in.  1.0  35.3 kip/in.    a in. 166 kip/in.  1.0  35.3 kip/in. 







Fu Ant  23.6 kip/in. t

Shear yielding component from AISC Manual Table 9-3b:



0.6Fy Agv t

 111 kip/in.

Shear rupture component from AISC Manual Table 9-3c:



Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

 83.7 kips  75.5 kips

ASD Tension rupture component from AISC Manual Table 9-3a:

0.6Fu Anv  125 kip/in. t

Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant +      a in. 125 kip/in.  1.0  23.6 kip/in.    a in. 111 kip/in.  1.0  23.6 kip/in.   55.7 kips  50.5 kips

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LRFD

ASD

Therefore:

Therefore:

Rn  75.5 kips  39.8 kips

o.k.

Rn  50.5 kips  26.5 kips o.k. 

Because the edge distance is smaller, block shear rupture is governed by the 3-in. leg. Flexural Yielding Strength of Angle The required flexural strength of the support leg of the angle is determined as follows: LRFD

ASD

M u  Ru e

M a  Ra e

0.300 in.     39.8 kips  1 w in.   2    75.6 kip-in.

0.300 in.     26.5 kips   1 w in.   2    50.4 kip-in.

The available flexural yielding strength of the support leg of the angle is determined as follows: M n  Fy Z x   a in.112 in.2     36 ksi   4    446 kip-in. LRFD

  0.90

  1.67

M n  0.90  446 kip-in.  401 kip-in.  75.6 kip-in. o.k.

ASD

M n 446 kip-in.   1.67  267 kip-in.  50.4 kip-in. o.k.

Flexural Rupture Strength of Angle The available flexural rupture strength of the support leg of the angle is determined as follows:  112 in.2  Z net   a in.   2 m in.  z in. 4.50 in.  2 m in.  z in.1.50 in.  4    8.46 in.3 M n  Fu Z net



  58 ksi  8.46 in.3

(Manual Eq. 9-4)



 491 kip-in.

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LRFD b  0.75 

 b  2.00 

 b M n  0.75  491 kip-in.



 368 kip-in.  75.6 kip-in. o.k.

ASD

M n 491 kip-in.  2.00 b  246 kip-in.  50.4 kip-in. o.k.

Flexural Yielding and Buckling of Coped Beam Web The required flexural strength of the coped section of the beam web is determined using AISC Manual Equation 95a or 9-5b, as follows: e  c  setback  4 in.  w in.  4.75 in.

LRFD 

M u  Ru e

ASD M a  Ra e

=  39.8 kips  4.75 in.

=  26.5 kips  4.75 in.

 189 kip-in.

 126 kip-in.

The minimum length of the connection elements is one-half of the reduced beam depth, ho: ho  d  d c (from AISC Manual Figure 9-2)  17.7 in.  2 in.  15.7 in.

 0.5ho

l

112 in.  0.5 15.7 in. 112 in.  7.85 in.

o.k.

The available flexural local buckling strength of a beam coped at the top flange is determined as follows: ho tw 15.7 in.  0.300 in.  52.3

 

(Manual Eq. 9-11)

c 4 in.  ho 15.7 in.  0.255

Because

c  1.0, the plate buckling coefficient, k, is calculated as follows: ho

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1.65

h  k  2.2  o   c 

(Manual Eq. 9-13a) 1.65

 15.7 in.   2.2    4 in.   21.0

c 4 in.  d 17.7 in.  0.226 Because

c  1.0, the buckling adjustment factor, f, is calculated as follows: d

c f  2  d  2  0.226 

(Manual Eq. 9-14a)

 0.452 k1  fk  1.61

(Manual Eq. 9-10)

  0.452  21.0   1.61  9.49  1.61  9.49 k1 E Fy

 p  0.475  0.475

(Manual Eq. 9-12)

 9.49  29, 000 ksi  50 ksi

 35.2

2 p  2  35.2   70.4 Because p <  ≤ 2p, calculate the nominal flexural strength using AISC Manual Equation 9-7. The plastic section modulus of the coped section, Znet, is determined from Table IV-11 (included in Part IV of this document).

Z net  32.1 in.3 M p  Fy Z net



  50 ksi  32.1 in.3



 1, 610 kip-in.

From AISC Manual Table 9-2: S net  18.2 in.3

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M y  Fy S net



  50 ksi  18.2 in.3



 910 kip-in.    M n  M p   M p  M y    1  p 

(Manual Eq. 9-7)

 52.3   1, 610 kip-in.  1, 610 kip-in.  910 kip-in.   1  35.2   1, 270 kip-in.

LRFD

ASD

b  0.90

 b  1.67

b M n  0.90 1, 270 kip-in.

M n 1, 270 kip-in.  b 1.67  760 kip-in.  126 kip-in. o.k.

 1,140 kip-in.  189 kip-in. o.k.

Shear Strength of Beam Web From AISC Specification Section J4.2(a), the available shear yielding strength of the beam web is determined as follows: Agv  ho tw  15.7 in. 0.300 in.  4.71 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  4.71 in.

2



 141 kips

LRFD

  1.50

  1.00 

 Rn  1.00 141 kips 

ASD

Rn 141 kips   1.50  94.0 kips  26.5 kips o.k.

 141 kips  39.8 kips o.k.

Block Shear Rupture of Beam Web The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the web is determined as follows, using AISC Manual Tables 9-3a, 93b and 9-3c, and AISC Specification Equation J4-5, with n = 4, lev = 12 in., leh = 12 in. (including a 4-in. tolerance to account for possible beam underrun), and Ubs = 1.0.

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LRFD Tension rupture component from AISC Manual Table 9-3a:  Fu Ant  51.8 kip/in.  t Shear yielding component from AISC Manual Table 9-3b:  0.60 Fy Agv  236 kip/in.   t

 Shear rupture component from AISC Manual Table 9-3c:



0.60Fu Anv  218 kip/in.  t

ASD Tension rupture component from AISC Manual Table 9-3a:





Fu Ant  34.5 kip/in. t

Shear yielding component from AISC Manual Table 9-3b:



0.60 Fy Agv t

 158 kip/in.

Shear rupture component from AISC Manual Table 9-3c:



 Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

0.60Fu Anv  145 kip/in. t

Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant  +     0.300 in. 145 kip/in.  1.0  34.5 kip/in. 

  0.300 in.  218 kip/in.  1.0  51.8 kip/in.    0.300 in.  236 kip/in.  1.0  51.8 kip/in.   80.9 kips  86.3 kips

  0.300 in. 158 kip/in.  1.0  34.5 kip/in.   53.9 kips  57.8 kips





Therefore:

Therefore:

Rn  80.9 kips  39.8 kips

o.k.

Rn  53.9 kips  26.5 kips o.k. 

Conclusion The connection is found to be adequate as given for the applied load.

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EXAMPLE II.A-28B ALL-BOLTED SINGLE ANGLE CONNECTION—STRUCTURAL INTEGRITY CHECK Given: Verify the all-bolted single-angle connection from Example II.A-28A, as shown in Figure II.A-28B-1, for the structural integrity provisions of AISC Specification Section B3.9. The connection is verified as a beam end connection. Note that these checks are necessary when design for structural integrity is required by the applicable building code. The angle is ASTM A36 material.

Fig. II.A-28B-1. Connection geometry for Example II.A-28B.

Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam and Girder ASTM A992 Fy = 50 ksi Fu = 65 ksi Angle ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W18x35

tw = 0.300 in.

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Girder W21x62

tw = 0.400 in. d = 21.0 in. kdes = 1.12 in. From AISC Specification Table J3.3, the hole diameter for w-in.-diameter bolts with standard holes is: dh = m in. From Example II.A-28A, the required shear strength is: LRFD

ASD

Vu  39.8 kips

Va  26.5 kips

From AISC Specification Section B3.9(b), the required axial tensile strength is: LRFD 2 Tu  Vu  10 kips 3 2   39.8 kips   10 kips 3  26.5 kips  10 kips

ASD Ta  Va  10 kips  26.5 kips  10 kips  26.5 kips

 26.5 kips Bolt Shear From AISC Specification Section J3.6, the nominal bolt shear strength is determined as follows: Fnv = 54 ksi, from AISC Specification Table J3.2

Tn  nFnv Ab



  4 bolts  54 ksi  0.442 in.2

(from Spec. Eq. J3-1)



 95.5 kips Bolt Tension From AISC Specification Section J3.6, the nominal bolt tensile strength is determined as follows: Fnt = 90 ksi, from AISC Specification Table J3.2

Tn  nFnt Ab



  4 bolts  90 ksi  0.442 in.

2

(from Spec. Eq. J3-1)



 159 kips Bolt Bearing and Tearout

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From AISC Specification Section B3.9, for the purpose of satisfying structural integrity requirements inelastic deformations of the connection are permitted; therefore, AISC Specification Equations J3-6b and J3-6d are used to determine the nominal bearing and tearout strength. For bolt bearing on the angle: Tn   4 bolts  3.0dtFu

(from Spec. Eq. J3-6b)

  4 bolts  3.0  w in. a in. 58 ksi   196 kips

For bolt bearing on the beam web: Tn   4 bolts  3.0dt w Fu

(from Spec. Eq. J3-6b)

  4 bolts  3.0  w in. 0.300 in. 65 ksi   176 kips

For bolt tearout on the angle:

lc  leh  0.5d h  12 in.  0.5 m in.  1.09 in. Tn   4 bolts 1.5lc tFu

(from Spec. Eq. J3-6d)

  4 bolts 1.5 1.09 in. a in. 58 ksi   142 kips

For bolt tearout on the beam web (including a 4-in. tolerance to account for possible beam underrun):

lc  leh  0.5d h  1w in.  4 in.  0.5 m in.  1.09 in. Tn   4 bolts 1.5lc tw Fu

(from Spec. Eq. J3-6d)

  4 bolts 1.5 1.09 in. 0.300 in. 65 ksi   128 kips

Angle Bending and Prying Action From AISC Manual Part 9, the nominal strength of the angle accounting for prying action is determined as follows:

t 2 a in.  1w in.  2  1.56 in.

b  gage 

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a  min 14 in., 1.25b  min 14 in., 1.25 1.56 in.  1.25 in. b  b 

db 2

 1.56 in. 

(Manual Eq. 9-18) w in. 2

 1.19 in.

d   d   a   a  b   1.25b  b  2   2   w in. w in.  1.25   1.25 1.56 in.  2 2  1.63 in.  2.33 in.  1.63 in.

b a 1.19 in.  1.63 in.  0.730



(Manual Eq. 9-23)

(Manual Eq. 9-22)

Note that end distances of 14 in. are used on the angles, so p is the average pitch of the bolts: l n 112 in.  4  2.88 in.

p

Check: p  s  3.00 in.

o.k.

d   dh m in.

d p m in.  1 2.88 in.  0.718

(Manual Eq. 9-20)

  1

Bn  Fnt Ab



  90 ksi  0.442 in.2



 39.8 kips/bolt

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4 Bn b pFu

tc 

(from Manual Eq. 9-26)

4  39.8 kips/bolt 1.19 in.



 2.88 in. 58 ksi 

 1.06 in.   tc  2  1    1  1     t    1.06 in. 2  1     1 0.718 1  0.730   a in.    5.63

 

(Manual Eq. 9-28)

Because    1 : 2

t  Q    1     tc  2

 a in.    1  0.718   1.06 in.   0.215 Tn   4 bolts  Bn Q

(from Manual Eq. 9-27)

  4 bolts  39.8 kips/bolt  0.215   34.2 kips

Block Shear Rupture—Angle From AISC Specification Section J4.3, the nominal block shear rupture strength of the angle with a “U” shaped failure plane is determined as follows: Tn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

where Agv  2leh t   2 12 in. a in.  1.13 in.2

Anv   2  leh  0.5  d h  z in.  t

  2  12 in.  0.5 m in.  z in.   a in.  0.797 in.2 Ant  9.00 in.  4  d h  z in.  t  9.00 in.  4 m in.  z in.   a in.  2.06 in.2

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U bs  1.0















Tn  0.60  58 ksi  0.797 in.2  1.0  58 ksi  2.06 in.2  0.60  36 ksi  1.13 in.2  1.0  58 ksi  2.06 in.2



 147 kips  144 kips Therefore: Tn  144 kips

Tensile Yielding of Angle From AISC Specification Section J4.1, the nominal tensile yielding strength of the angle is determined as follows: Ag  lt  112 in. a in.  4.31 in.2

Tn  Fy Ag

(from Spec. Eq. J4-1)



  36 ksi  4.31 in.

2



 155 kips

Tensile Rupture of Angle From AISC Specification Section J4.1, the nominal tensile rupture strength of the angle is determined as follows: Ae  AnU

(Spec. Eq. D3-1)

 l  n  d h  z in.  tU  112 in.  4 m in.  z in.   a in.1.0   3.00 in.2

Tn  Fu Ae



  58 ksi  3.00 in.

2

(from Spec. Eq. J4-2)



 174 kips Block Shear Rupture—Beam Web From AISC Specification Section J4.3, the nominal block shear rupture strength of the beam web with a “U” shaped failure plane is determined as follows (including a 4-in. tolerance to account for possible beam underrun): Tn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

where Agv  2leh tw   2 1w in.  4 in. 0.300 in.  0.900 in.2

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Anv   2  leh  0.5  d h  z in.  tw

  2  1w in.  4 in.  0.5 m in.  z in.   0.300 in.  0.638 in.2 Ant  9.00 in.  3  d h  z in.  t w  9.00 in.  3 m in.  z in.   0.300 in.  1.91 in.2 U bs  1.0















Tn  0.60  65 ksi  0.638 in.2  1.0  65 ksi  1.91 in.2  0.60  50 ksi  0.900 in.2  1.0  65 ksi  1.91 in.2  149 kips  151 kips Therefore: Tn  149 kips

Nominal Tensile Strength The controlling tensile strength, Tn, is the least of those previously calculated: 95.5 kips, 159 kips, 196 kips, 176 kips, 142 kips, 128 kips, 34.2 kips, 144 kips, 155 kips,  Tn  min   174 kips, 149 kips   34.2 kips

LRFD Tn  34.2 kips  26.5 kips o.k.

ASD Tn  34.2 kips  26.5 kips o.k.

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EXAMPLE II.A-29 FLANGE)

BOLTED/WELDED SINGLE-ANGLE CONNECTION (BEAM-TO-COLUMN

Given: Verify a single-angle connection between an ASTM A992 W1650 beam and an ASTM A992 W1490 column flange, as shown in Figure II.A-29-1, to support the following beam end reactions: RD = 9 kips RL = 27 kips Use an ASTM A36 single angle. Use 70-ksi electrode welds to connect the single angle to the column flange.

Fig. II.A-29-1. Connection geometry for Example II.A-29.

Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Angle ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1650 tw = 0.380 in. d = 16.3 in. tf = 0.630 in.

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Column W1490 tf = 0.710 From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  9 kips   1.6  27 kips 

ASD Ra  9 kips  27 kips  36.0 kips

 54.0 kips Single Angle, Bolts and Welds Check eccentricity of the connection. For the 4-in. angle leg attached to the supported beam:

e = 2w in. < 3.00 in., therefore, eccentricity does not need to be considered for this leg. For the 3-in. angle leg attached to the supporting column flange: Because the half-web dimension of the W1650 supported beam is less than 4 in., AISC Manual Table 10-12 may conservatively be used. Use a four-bolt single-angle (L43a). From AISC Manual Table 10-12, the bolt and angle available strength is: LRFD Rn  71.4 kips  54.0 kips

o.k.

ASD

Rn  47.6 kips  36.0 kips o.k. 

From AISC Manual Table 10-12, the available weld strength for a x-in. fillet weld is: LRFD Rn  56.6 kips  54.0 kips

o.k.

ASD

Rn  37.8 kips  36.0 kips o.k. 

Support Thickness The minimum support thickness that matches the column flange strength to the x-in. fillet weld strength is: tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  3 

65 ksi  0.143 in.  0.710 in.

o.k.

Note: The minimum thickness values listed in Table 10-12 are for conditions with angles on both sides of the web. Use a four-bolt single-angle, L43a. The 3-in. leg will be shop welded to the column flange and the 4-in. leg will be field bolted to the beam web.

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Supported Beam Web The available bearing and tearout strength of the beam web is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

Rn   4 bolts  87.8 kip/in. 0.380 in.  133 kips  54.0 kips o.k.

ASD Rn   4 bolts  58.5 kip/in. 0.380 in.   88.9 kips  36.0 kips o.k.

Conclusion The connection is found to be adequate as given for the applied load.

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EXAMPLE II.A-30 ALL-BOLTED TEE CONNECTION (BEAM-TO-COLUMN FLANGE) Given: Verify an all-bolted tee connection between an ASTM A992 W1650 beam and an ASTM A992 W1490 column flange, as shown in Figure II.A-30-1, to support the following beam end reactions: RD = 9 kips RL = 27 kips Use an ASTM A992 WT522.5 with a four-bolt connection.

Fig. II.A-30-1. Connection geometry for Example II.A-30.

Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam, column and tee ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Tables 1-1 and 1-8, the geometric properties are as follows: Beam W1650 tw = 0.380 in. d = 16.3 in. tf = 0.630 in. Column W1490 tf = 0.710 in.

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Tee WT522.5

d bf tf tsw k1 kdes

= 5.05 in. = 8.02 in. = 0.620 in. = 0.350 in. = m in. (see W1045 AISC Manual Table 1-1) = 1.12 in.

From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  9 kips   1.6  27 kips 

ASD Ra  9 kips  27 kips  36.0 kips

 54.0 kips Limitation on Tee Stem or Beam Web Thickness

See rotational ductility discussion at the beginning of the AISC Manual Part 9. For the tee stem, the maximum tee stem thickness is: d  z in. 2 w in.   z in. 2  0.438 in.  0.350 in. o.k.

tsw max 

(Manual Eq. 9-39)

For W1650 beam web, the maximum beam web thickness is: d  z in. 2 w in.   z in. 2  0.438 in.  0.380 in. o.k.

t w max 

(from Manual Eq. 9-39)

Limitation on Bolt Diameter for Bolts through Tee Flange Note: The bolts are not located symmetrically with respect to the centerline of the tee. b  flexible width in connection element (see AISC Manual Figure 9-6) t t  2w in.  sw  w  k1 2 2 0.350 in. 0.380 in.  2w in.    m in. 2 2  1.57 in.

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d min  0.163t f

 Fy  b 2  2  2   0.69 tsw b l 

(Manual Eq. 9-38)

2   50 ksi   1.57 in.    0.69 0.350 in. 2  0.163  0.620 in.     1.57 in.   112 in.2   0.810 in.  0.408 in.

Therefore: d min  0.408 in.  w in. o.k.

Because the connection is rigid at the support, the bolts through the tee stem must be designed for shear, but do not need to be designed for an eccentric moment. Strength of the Bolted Connection—Tee From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the invividual strengths of the individual fasteners, taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is: LRFD

ASD

rn  11.9 kips/bolt 

rn  17.9 kips/bolt

The available bearing and tearout strength of the tee at the bottom edge bolt is determined using AISC Manual Table 7-5, with le = 14 in., as follows: LRFD rn   49.4 kip/in. 0.350 in.  17.3 kips/bolt

ASD rn   32.9 kip/in. 0.350 in.   11.5 kips/bolt

The bearing or tearout strength controls over bolt shear for the bottom edge bolt in the tee. The available bearing and tearout strength of the tee at the interior bolts (not adjacent to the edge) is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

rn   87.8 kip/in. 0.350 in.  30.7 kips/bolt

ASD rn   58.5 kip/in. 0.350 in.   20.5 kips/bolt

The bolt shear strength controls over bearing or tearout for the interior bolts in the tee. The strength of the bolt group in the beam web is determined by summing the strength of the individual fasteners as follows:

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LRFD

ASD Rn  1 bolt 11.5 kips/bolt     3 bolts 11.9 kips/bolt 

Rn  1 bolt 17.3 kips/bolt    3 bolts 17.9 kips/bolt   71.0 kips  54.0 kips

o.k.

 47.2 kips  36.0 kips o.k.

Strength of the Bolted Connection—Beam Web The available bearing and tearout strength for all bolts in the beam web is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

ASD rn   58.5 kip/in. 0.380 in.   22.2 kips/bolt

rn   87.8 kip/in. 0.380 in.  33.4 kips/bolt

The bolt shear strength controls over bearing or tearout in the beam web; therefore, the beam web is adequate based on previous calculations. Flexural Yielding of Stem The flexural yielding strength is checked at the junction of the stem and the fillet. The required flexural strength is determined as follows: LRFD

ASD

M u  Pu e

M a  Pa e

 Pu  a  kdes 

 Pa  a  kdes 

  54.0 kips  3.80 in.  1.12 in.

  36.0 kips  3.80 in.  1.12 in.

 145 kip-in.

 96.5 kip-in.

The available flexural strength of the tee stem is determined as follows:   0.90

LRFD

 M n  Fy Z x

  0.350 in.112 in.2    0.90  50 ksi   4    521 kip-in. > 145 kip-in. o.k.

  1.67  M n Fy Z x   

ASD

2  50 ksi    0.350 in.112 in.      4  1.67     346 kip-in. > 96.5 kip-in. o.k.

Shear Strength of Stem From AISC Specification Section J4.2(a), the available shear yielding strength of the tee stem is determined as follows: Agv  ltsw  112 in. 0.350 in.  4.03 in.2

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Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  4.03 in.

2



 121 kips

LRFD

  1.00

  1.50

Rn  1.00 121 kips 

ASD

Rn 121 kips   1.50  80.7 kips  36.0 kips o.k.

 121 kips  54.0 kips o.k.

From AISC Specification Section J4.2, the available shear rupture strength of the tee stem is determined using the net area determined in accordance with AISC Specification Section B4.3b.

Anv  l  n  d h  z in.  t sw  112 in.  4 m in.  z in.   0.350 in.  2.80 in.2

Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  65 ksi  2.80 in.2



 109 kips

  0.75

LRFD

  2.00

Rn  0.75 109 kips 

ASD

Rn 109 kips   2.00  54.5 kips  36.0 kips o.k.

 81.8 kips  54.0 kips o.k.

Block Shear Rupture of Stem The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the tee stem is determined as follows, using AISC Manual Tables 9-3a, 9-3b and 9-3c, and AISC Specification Equation J4-5, with n = 4, leh = lev = 14 in., and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a:  Fu Ant  39.6 kip/in.   t

ASD Tension rupture component from AISC Manual Table 9-3a:





Fu Ant  26.4 kip/in.  t

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LRFD Shear yielding component from AISC Manual Table 9-3b:  0.60 Fy Agv  231 kip/in.  t

ASD Shear yielding component from AISC Manual Table 9-3b:

Shear rupture component from AISC Manual Table 9-3c:

Shear rupture component from AISC Manual Table 9-3c:





0.60Fu Anv  210 kip/in.  t





The design block shear rupture strength is:  Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

 154 kip/in.

0.60Fu Anv  140 kip/in. t

The allowable block shear rupture strength is: Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant +      0.350 in. 140 kip/in.  1.0  26.4 kip/in. 

  0.350 in.  210 kip/in.  1.0  39.6 kip/in.    0.350 in.  231 kip/in.  1.0  39.6 kip/in.   87.4 kips  94.7 kips

  0.350 in. 154 kip/in.  1.0  26.4 kip/in.   58.2 kips  63.1 kips

  Therefore: Rn  87.4 kips  54.0 kips

t







0.60 Fy Agv

 Therefore: o.k.

Rn  58.2 kips  36.0 kips o.k. 

Because the connection is rigid at the support, the bolts attaching the tee flange to the support must be designed for the shear and the eccentric moment. Bolt Group at Column Check bolts for shear and bearing combined with tension due to eccentricity. The following calculation follows the Case II approach in the Section “Eccentricity Normal to the Plane of the Faying Surface” in Part 7 of the AISC Manual. The available shear strength of the bolts is determined as follows: LRFD rn  17.9 kips/bolt (from AISC Manual Table 7-1)

Pu (Manual Eq. 7-13a) n 54.0 kips  8 bolts  6.75 kips/bolt  17.9 kips/bolt o.k.

ASD

rn  11.9 kips/bolt (from AISC Manual Table 7-1)  Pa (Manual Eq. 7-13b) n 36.0 kips  8 bolts  4.50 kips/bolt  11.9 kips/bolt o.k.

ruv 

rav 

Ab  0.442 in.2 (from AISC Manual Table 7-1)

Ab  0.442 in.2 (from AISC Manual Table 7-1)

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LRFD f rv

ASD

r  uv Ab 6.75 kips/bolt  0.442 in.2  15.3 ksi

r f rv  av Ab 4.50 kips/bolt  0.442 in.2  10.2 ksi

The nominal tensile stress modified to include the effects of shear stress is determined from AISC Specification Section J3.7 as follows. From AISC Specification Table J3.2: Fnt  90 ksi Fnv  54 ksi LRFD Tensile force per bolt, rut: rut  

ASD Tensile force per bolt, rat:

Pu e nd m

(Manual Eq. 7-14a)

 54.0 kips  3.80 in.  4 bolts  6.00 in.



 8.55 kips/bolt

Pa e nd m

(Manual Eq. 7-14b)

 36.0 kips  3.80 in.  4 bolts  6.00 in.

 5.70 kips/bolt   2.00 

  0.75

Fnt f rv  Fnt (Spec. Eq. J3-3a) Fnv 90 ksi  1.3  90 ksi   15.3 ksi   90 ksi 0.75  54 ksi 

Fnt  1.3Fnt 

 83.0 ksi  90 ksi  83.0 ksi rn  Fnt Ab

rat 



 0.75  83.0 ksi  0.442 in.2

(from Spec. Eq. J3-2)



 27.5 kips/bolt  8.55 kips/bolt o.k.

Fnt  1.3Fnt 

Fnt f rv  Fnt Fnv

 1.3  90 ksi  

2.00  90 ksi 

54 ksi  83.0 ksi  90 ksi  83.0 ksi

rn Fnt Ab    

(Spec. Eq. J3-3b)

10.2 ksi   90 ksi

(from Spec. Eq. J3-2)

83.0 ksi   0.442 in.2 

2.00  18.3 kips/bolt  5.70 kips/bolt

o.k.

With le = 14 in. and s = 3 in., the bearing or tearout strength of the tee flange exceeds the single shear strength of the bolts. Therefore, the bearing and tearout strength is adequate. Prying Action From AISC Manual Part 9, the available tensile strength of the bolts taking prying action into account is determined as follows. By inspection, prying action in the tee will control over prying action in the column. Note: The bolts are not located symmetrically with respect to the centerline of the tee.

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bf

tw tsw   2w in. 2 2 2 8.02 in. 0.380 in. 0.350 in.     2w in. 2 2 2  0.895 in.

a



tw 2 0.380 in.  2w in.  2  2.94 in.

b  2w in. 

d  a   a  b 2 

db      1.25b   2    w in. w in.  0.895 in.   1.25  2.94 in.  2 2  1.27 in.  4.05 in.  1.27 in.

(Manual Eq. 9-23)

d   b   b  b  2  

(Manual Eq. 9-18)

 2.94 in. 

w in. 2

 2.57 in.

 

b a

(Manual Eq. 9-22)

2.57 in. 1.27 in.

 2.02

p  lev  0.5s  14 in.  0.5  3 in.  2.75 in. Check: p s 2.75 in.  3 in.

o.k.

 lev  1.75b

p

2.75 in.  14 in.  1.75  2.94 in. 2.75 in.  6.40 in.

o.k.

d   dh  m in.

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d p m in.  1 2.75 in.  0.705

  1

(Manual Eq. 9-20)

LRFD

ASD

Tr  rut

Tr  rat

 8.55 kips/bolt

 5.70 kips/bolt

Bc  rn

rn   18.3 kips/bolt

Bc 

 27.5 kips/bolt

 

1  Bc   1   Tr 

(Manual Eq. 9-21)

1  27.5 kips/bolt      1 2.02  8.55 kips/bolt  

 

 1.10

1  Bc   1   Tr  1  18.3 kips/bolt      1 2.02  5.70 kips/bolt  

 1.09

Because   1 , set    1.0.

Because   1 , set    1.0.

  0.90

  1.67

tmin 



(Manual Eq. 9-21)

4Tu b pFu 1   

(Manual Eq. 9-19a)

4  8.55 kips/bolt  2.57 in.

0.90  2.75 in. 65 ksi  1   0.705 1.0  

tmin 



 4Ta b pFu 1    

(Manual Eq. 9-19b)

1.67  4  5.70 kips/bolt  2.57 in.

 2.75 in. 65 ksi  1   0.705 1.0 

 0.567 in.  0.620 in. o.k.

 0.566 in.  0.620 in. o.k.

Similarly, checks of the tee flange for shear yielding, shear rupture, and block shear rupture will show that the tee flange is adequate. Bolt Bearing on Column Flange The available bearing and tearout strength of the column flange is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

Rn  8 bolts  87.8 kip/in. 0.710 in.  499 kips  54.0 kips o.k.

ASD Rn   8 bolts  58.5 kip/in. 0.710 in.   332 kips  36.0 kips o.k.

Note: Although the edge distance (a = 0.895 in.) for one row of bolts in the tee flange does not meet the minimum value indicated in AISC Specification Table J3.4, based on footnote [a], the edge distance provided is acceptable because the provisions of AISC Specification Section J3.10 and J4.4 have been met in this case.

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Conclusion The connection is found to be adequate as given for the applied load.

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EXAMPLE II.A-31 BOLTED/WELDED TEE CONNECTION (BEAM-TO-COLUMN FLANGE) Given:

Verify the tee connection bolted to an ASTM A992 W1650 supported beam and welded to an ASTM A992 W1490 supporting column flange, as shown in Figure II.A-31-1, to support the following beam end reactions: RD = 6 kips RL = 18 kips Use 70-ksi electrodes. Use an ASTM A992 WT522.5 with a four-bolt connection to the beam web.

Fig. II.A-31-1. Connection geometry for Example II.A-31. Solution:

From AISC Manual Table 2-4, the material properties are as follows: Beam, column and tee ASTM A992 Fy = 50 ksi Fu = 65 ksi From AISC Manual Tables 1-1 and 1-8, the geometric properties are as follows: Beam W1650 tw = 0.380 in. d = 16.3 in. tf = 0.630 in. Column W1490 tf = 0.710 in.

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Tee WT522.5

d bf tf tsw k1 kdes

= 5.05 in. = 8.02 in. = 0.620 in. = 0.350 in. = m in. (see W1045, AISC Manual Table 1-1) = 1.12 in.

From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  6 kips   1.6 18 kips 

ASD Ra  6 kips  18 kips  24.0 kips

 36.0 kips Limitation on Tee Stem or Beam Web Thickness

See rotational ductility discussion at the beginning of AISC Manual Part 9. For the tee stem, the maximum tee stem thickness is: d  z in. 2 w in.   z in. 2  0.438 in.  0.350 in. o.k.

tsw max 

(Manual Eq. 9-39)

For W1650 beam web, the maximum beam web thickness is: d  z in. 2 w in.   z in. 2  0.438 in.  0.380 in. o.k.

t w max 

(Manual Eq. 9-39)

Weld Design b  flexible width in connection element b f  2k1  2 8.02 in.  2 m in.  2  3.20 in.

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Fy t 2f b

 b2   l 2  2    s  tsw     50 ksi  0.620 in.2    3.20 in.2    0.0155   2    s  0.350 in. 2 3.20 in.    112 in.   0.193 in.  0.219 in.  0.193 in.

wmin  0.0155

(Manual Eq. 9-37)

The minimum weld size is 4 in. per AISC Specification Table J2.4. Try 4-in. fillet welds. From AISC Manual Table 10-2, with n = 4, l = 112 in., and Welds B = 4 in.: LRFD Rn  79.9 kips  36.0 kips

o.k. 

ASD

Rn  53.3 kips  24.0 kips o.k. . 

Use 4-in. fillet welds. Supporting Column Flange From AISC Manual Table 10-2, with n = 4, l = 112 in., and Welds B = 4 in., the minimum support thickness is 0.190 in.

t f  0.710 in.  0.190 in. o.k. Strength of Bolted Connection From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the invividual strengths of the individual fasteners, taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. The 3-in. angle leg attached to the supporting girder will control because eccentricity must be taken into consideration. Because the connection is flexible at the support, the tee stem and bolts must be designed for eccentric shear, where the eccentricity, eb, is determined as follows:

eb  a  d  leh  5.05 in.  14 in.  3.80 in. From AISC Manual Table 7-6 for Angle = 00, with s = 3 in., ex = eb = 3.80 in., and n = 4: C  2.45

From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is:

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LRFD

ASD

rn  11.9 kips/bolt 

rn  17.9 kips/bolt

The available bearing and tearout strength of the tee at the bottom edge bolt is determined using AISC Manual Table 7-5, with le = 14 in., as follows: LRFD

rn   49.4 kip/in. 0.350 in.  17.3 kips/bolt

ASD rn   32.9 kip/in. 0.350 in.   11.5 kips/bolt

The available bearing and tearout strength of the tee at the interior bolts (not adjacent to the edge) is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

rn   87.8 kip/in. 0.350 in.  30.7 kips/bolt

ASD rn   58.5 kip/in. 0.350 in.   20.5 kips/bolt

Note: By inspection, bolt bearing on the beam web does not control. The available strength of the bolted connection is determined from AISC Manual Equation 7-16, conservatively using the minimum available strength calculated for bolt shear, bearing on the tee, and tearout on the tee. LRFD

Rn  C rn  2.45 17.3 kips/bolt   42.4 kips  36.0 kips o.k.

ASD Rn rn C    2.45 11.5 kips/bolt   28.2 kips  24.0 kips o.k.

Flexural Yielding of Tee Stem The required flexural strength of the tee stem is determined as follows: LRFD

M u  Pu eb

ASD

M a  Pa eb

  36.0 kips  3.80 in.

  24.0 kips  3.80 in.

 137 kip-in.

 91.2 kip-in.

The available flexural yielding strength of the tee stem is determined as follows:

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LRFD

  0.90 

 M n  Fy Z x   0.350 in.112 in.2    0.90  50 ksi   4    521 kip-in.  137 kip-in. o.k.

  1.67   M n Fy Z x   

ASD

2 50 ksi   0.350 in.112 in.    1.67  4    346 kip-in.  91.2 kip-in. o.k.



Flexural Rupture of Tee Stem The available flexural rupture strength of the plate is determined as follows: Z net

 112 in.2    0.350 in.   2 m in.  z in. 4.50 in.  2 m in.  z in.1.50 in.  4    7.90 in.3

M n  Fu Z net

(Manual Eq. 9-4)



  65 ksi  7.90 in.

3



 514 kip-in.

LRFD

  0.75

M n  0.75  514 kip-in.  386 kip-in.  137 kip-in. o.k.

ASD   2.00   M n 514 kip-in.   2.00  257 kip-in.  91.2 kip-in. o.k.

Shear Strength of Stem From AISC Specification Section J4.2(a), the available shear yielding strength of the tee stem is determined as follows: Agv  ltsw  112 in. 0.350 in.  4.03 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  4.03 in.2



 121 kips

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LRFD

  1.00

ASD

  1.50

Rn  1.00 121 kips 

Rn 121 kips   1.50  80.7 kips  24.0 kips o.k.

 121 kips  36.0 kips o.k.

From AISC Specification Section J4.2(b), the available shear rupture strength of the tee stem is determined using the net area determined in accordance with AISC Specification Section B4.3b.

Anv  l  n  d h  z in.  t sw  112 in.  4 m in.  z in.   0.350 in.  2.80 in.2

Rn  0.60 Fu Anv

(Spec. Eq. J4-4)



 0.60  65 ksi  2.80 in.

2



 109 kips

  0.75

LRFD

ASD

  2.00

Rn  0.75 109 kips 

Rn 109 kips  2.00   54.5 kips  24.0 kips o.k.

 81.8 kips  36.0 kips o.k. Block Shear Rupture of Stem

The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the tee stem is determined as follows, using AISC Manual Tables 9-3a, 9-3b and 9-3c, and AISC Specification Equation J4-5, with n = 4, leh = lev = 14 in., and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a:  Fu Ant  39.6 kip/in.  t Shear yielding component from AISC Manual Table 9-3b:  0.60 Fy Agv  231 kip/in.   t

ASD Tension rupture component from AISC Manual Table 9-3a:





Fu Ant  26.4 kip/in. t

Shear yielding component from AISC Manual Table 9-3b:





0.60 Fy Agv t

 154 kip/in. 

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LRFD Shear rupture component from AISC Manual Table 9-3c:



0.60Fu Anv  210 kip/in.   t  The design block shear rupture strength is:  Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

ASD Shear rupture component from AISC Manual Table 9-3c:





0.60Fu Anv  140 kip/in.  t

 The allowable block shear rupture strength is: Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant  +     0.350 in. 140 kip/in.  1.0  26.4 kip/in. 

  0.350 in.  210 kip/in.  1.0  39.6 kip/in.    0.350 in.  231 kip/in.  1.0  39.6 kip/in.   87.4 kips  94.7 kips

 

  0.350 in. 154 kip/in.  1.0  26.4 kip/in.   58.2 kips  63.1 kips

 Therefore: Rn  87.4 kips  36.0 kips

Therefore: o.k.

Rn  58.2 kips  24.0 kips o.k. 

Conclusion The connection is found to be adequate as given for the applied load.

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Chapter IIB Fully Restrained (FR) Moment Connections The design of fully restrained (FR) moment connections is covered in Part 12 of the AISC Manual.

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EXAMPLE II.B-1 BOLTED FLANGE-PLATED FR MOMENT CONNECTION (BEAM-TO-COLUMN FLANGE) Given: Verify a bolted flange-plated FR moment connection between an ASTM A992 W1850 beam and an ASTM A992 W1499 column flange, as shown in Figure II.B-1-1, to transfer the following beam end reactions: Vertical shear: VD = 7 kips VL = 21 kips Strong-axis moment: MD = 42 kip-ft ML = 126 kip-ft Use 70-ksi electrodes. The flange and web plates are ASTM A36 material. Check the column for stiffening requirements.

Fig. II.B-1-1. Connection geometry for Example II.B-1.

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Solution: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Plates ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1850

d = 18.0 in. bf = 7.50 in. tf = 0.570 in. tw = 0.355 in. Sx = 88.9 in.3 Column W1499

d = 14.2 in. bf = 14.6 in. tf = 0.780 in. From AISC Specification Table J3.3, the hole diameter for a d-in.-diameter bolt with standard holes is: d h  , in.

From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  7 kips   1.6  21 kips   42.0 kips

M u  1.2  42 kip-ft   1.6 126 kip-ft   252 kip-ft

ASD

Ra  7 kips  21 kips  28.0 kips M a  42 kip-ft  126 kip-ft  168 kip-ft

Flexural Strength of Beam From AISC Specification Section F13.1, the available flexural strength of the beam is limited according to the limit state of tensile rupture of the tension flange. A fg  b f t f   7.50 in. 0.570 in.  4.28 in.2

The net area of the flange is determined in accordance with AISC Specification Section B4.3b.

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A fn  A fg   2 bolts  d h  z in. t f  4.28 in.2   2 bolts , in.  z in. 0.570 in.  3.14 in.2 Fy

50 ksi Fu 65 ksi  0.769  0.8; therefore, Yt  1.0 



Fu A fn   65 ksi  3.14 in.2



 204 kips



Yt Fy A fg  1.0  50 ksi  4.28 in.2



 214 kips  204 kips

Therefore, the nominal flexural strength, Mn, at the location of the holes in the tension flange is not greater than:

Mn 

Fu A fn A fg

Sx

(Spec. Eq. F13-1)





 204 kips  88.9 in.3  2   4.28 in.   4, 240 kip-in. or 353 kip-ft LRFD

ASD

b  0.90 

b  1.67 

M n  0.90  353 kip-ft 

M n 353 kip-ft  b 1.67  211 kip-ft  168 kip-ft o.k.

 318 kip-ft  252 kip-ft o.k.

Note: The available flexural strength of the beam may be less than that determined based on AISC Specification Equation F13-1. Other applicable provisions in AISC Specification Chapter F should be checked to possibly determine a lower value for the available flexural strength of the beam. Single-Plate Web Connection Strength of the bolted connection—web plate From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the strengths of the individual fasteners, which may be taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. From AISC Manual Table 7-1, the available shear strength per bolt for d-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is:

LRFD

ASD

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rn  16.2 kips/bolt 

rn  24.3 kips/bolt

The available bearing strength of the plate per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration: rn  2.4dtFu

(Spec. Eq. J3-6a)

 2.4  d in. a in. 58 ksi   45.7 kips/bolt

  0.75

LRFD

rn  0.75  45.7 kips/bolt   34.3 kips/bolt

  2.00

ASD

rn 45.7 kips/bolt  2.00   22.9 kips/bolt

The available tearout strength of the plate at the interior bolts is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration.

lc  s  d h  3 in.  , in.  2.06 in. rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2  2.06 in. a in. 58 ksi   53.8 kips/bolt  = 0.75

LRFD

rn = 0.75  53.8 kips/bolt   40.4 kips/bolt

  2.00

ASD

rn 53.8 kips/bolt   2.00  26.9 kips/bolt

Therefore, bolt shear controls over bearing or tearout at interior bolts. The available tearout strength of the plate at the edge bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration.

lc  lev  0.5  d h   12 in.  0.5 , in.  1.03 in. rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2 1.03 in. a in. 58 ksi   26.9 kips/bolt LRFD

ASD

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 = 0.75

  2.00

rn = 0.75  26.9 kips/bolt 

rn 26.9 kips/bolt   2.00  13.5 kips/bolt

 20.2 kips/bolt

Therefore, tearout controls over bolt shear or bearing at the edge bolt. The strength of the bolt group in the plate is determined by summing the strength of the individual fasteners as follows: LRFD Rn  1 bolt  20.2 kips/bolt 

ASD

Rn 

  2 bolts  24.3 kips/bolt   68.8 kips  42.0 kips o.k.

 1 bolt 13.5 kips/bolt    2 bolts 16.2 kips/bolt   45.9 kips  28.0 kips o.k.

Strength of the bolted connection—beam web

Because there are no edge bolts, the available bearing and tearout strength of the beam web for all bolts is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

rn  102 kip/in. 0.355 in.  36.2 kips/bolt

ASD rn   68.3 kip/in. 0.355 in.   24.2 kips/bolt

Bolt shear strength is the governing limit state for all bolts at the beam web. The strength of the bolt group in the beam web is determined by summing the strength of the individual fasteners as follows: LRFD Rn   3 bolts  24.3 kips/bolt   72.9 kips  42.0 kips

o.k.

ASD

Rn 

  3 bolts 16.2 kips/bolt   48.6 kips  28.0 kips o.k.

Shear strength of the web plate

From AISC Specification Section J4.2(a), the available shear yielding strength of the plate is determined as follows:

Agv  lt   9 in. a in.  3.38 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  3.38 in.2



 73.0 kips LRFD

ASD

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  1.00

  1.50

Rn  1.00  73.0 kips 

Rn 73.0 kips   1.50  48.7 kips  28.0 kips

 73.0 kips  42.0 kips

o.k.

o.k.

From AISC Specification Section J4.2(b), the available shear rupture strength of the plate is determined as follows: Anv  l  n  d h  z in.  t  9 in.   3 bolts , in.  z in.   a in.  2.25 in.2 Rn  0.60 Fu Anv



 0.60  58 ksi  2.25 in.

2

(Spec. Eq. J4-4)



 78.3 kips

  0.75

LRFD

  2.00

Rn  0.75  78.3 kips   58.7 kips  42.0 kips

ASD

Rn 78.3 kips   2.00  39.2 kips  28.0 kips o.k.

o.k.

Block shear rupture of the web plate The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60Fu Anv  Ubs Fu Ant  0.60Fy Agv  Ubs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the web plate is determined as follows, using AISC Manual Tables 93a, 9-3b and 9-3c, and AISC Specification Equation J4-5, with n = 3, leh = 2 in., lev = 12 in., and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a:  Fu Ant  65.3 kip/in.  t Shear yielding component from AISC Manual Table 9-3b:  0.60 Fy Agv  121 kip/in.  t

ASD Tension rupture component from AISC Manual Table 9-3a:





Fu Ant  43.5 kip/in. t

Shear yielding component from AISC Manual Table 9-3b:





0.60Fy Agv  81.0 kip/in. t

LRFD

ASD

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Shear rupture component from AISC Manual Table 9-3c:

Shear rupture component from AISC Manual Table 9-3c:

0.60Fu Anv  131 kip/in.   t





The design block shear rupture strength is: Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant 

  a in. 131 kip/in.  1.0  65.3 kip/in.    a in. 121 kip/in.  1.0  65.3 kip/in.   73.6 kips  69.9 kips



0.60Fu Anv  87.0 kip/in.  t

 he allowable block shear rupture strength is:

Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant  +     a in. 87.0 kip/in.  1.0  43.5 kip/in.    a in. 81.0 kip/in.  1.0  43.5 kip/in.   48.9 kips  46.7 kips

Therefore:

Therefore:

Rn  69.9 kips  42.0 kips

o.k.

Rn  46.7 kips  28.0 kips 

o.k.

Weld shear strength of the web plate to the column flange The available weld strength is determined using AISC Manual Equations 8-2a or 8-2b, with the assumption that the weld is in direct shear (the incidental moment in the weld plate due to eccentricity is absorbed by the flange plates). D  4 (for a 4 -in. fillet weld)

LRFD Rn   2 welds 1.392 kip/in. Dl

ASD Rn   2 welds  0.928 kip/in. Dl

  2 welds 1.392 kip/in. 4  9 in. 

  2 welds  0.928 kip/in. 4  9 in. 

 100 kips  42.0 kips

 66.8 kips  28.0 kips

o.k.

o.k.

Column flange rupture strength at welds From AISC Specification Section J4.2(b), the available shear rupture strength of the column flange is determined as follows: Anv   2 welds  lt f

  2 welds  9 in. 0.780 in.  14.0 in.2 Rn  0.60 Fu Anv



 0.60  65 ksi  14.0 in.

2

(Spec. Eq. J4-4)



 546 kips

LRFD

ASD

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  0.75

  2.00

Rn  0.75  546 kips 

Rn 546 kips   2.00  273 kips  28.0 kips o.k.

 410 kips  42.0 kips

o.k.

Flange Plate Connection Flange force

The moment arm between flange forces, dm, used for verifying the fastener strength is equal to the depth of the beam. This dimension represents the faying surface between the flange of the beam and the tension plate. LRFD Puf  

Mu dm

ASD (Manual Eq. 12-1a)

 252 kip-ft 12 in./ft 

Paf  

18.0 in.  168 kips

Ma dm

(Manual Eq. 12-1b)

168 kip-ft 12 in./ft 

18.0 in.  112 kips

Strength of the bolted connection—flange plate

From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the strengths of the individual fasteners, which may be taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. From AISC Manual Table 7-1, the available shear strength per bolt for d-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is: LRFD

ASD rn  16.2 kips/bolt 

rn  24.3 kips/bolt

The available bearing strength of the plate per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration: rn  2.4dtFu

(Spec. Eq. J3-6a)

 2.4  d in. w in. 58 ksi   91.4 kips/bolt

  0.75

LRFD

rn  0.75  91.4 kips/bolt   68.6 kips/bolt

  2.00

ASD

rn 91.4 kips/bolt   2.00  45.7 kips/bolt

The available tearout strength of the plate at the interior bolts is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration.

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lc  s  d h

 3 in.  , in.  2.06 in. rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2  2.06 in. w in. 58 ksi   108 kips/bolt  = 0.75

LRFD

  2.00

rn = 0.75 108 kips/bolt 

ASD

rn 108 kips/bolt   2.00  54.0 kips/bolt

 81.0 kips/bolt

Therefore, bolt shear controls over bearing or tearout at interior bolts. The available tearout strength of the plate at the edge bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration.

lc  lev  0.5  d h   12 in.  0.5 , in.  1.03 in. rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2 1.03 in. w in. 58 ksi   53.8 kips/bolt  = 0.75

LRFD

  2.00

rn = 0.75  53.8 kips/bolt 

ASD

rn 53.8 kips/bolt   2.00  26.9 kips/bolt

 40.4 kips/bolt

Therefore, bolt shear controls over bearing or tearout at edge bolts. The strength of the bolt group in the beam web is determined by summing the strength of the individual fasteners as follows: LRFD Rn   8 bolts  24.3 kips/bolt   194 kips  168 kips

ASD

Rn 

o.k.

  8 bolts 16.2 kips/bolt   130 kips  112 kips o.k.

Strength of the bolted connection—beam flange

The available bearing strength of the flange per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration:

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rn  2.4dtFu

(Spec. Eq. J3-6a)

 2.4  d in. 0.570 in. 65 ksi   77.8 kips/bolt

  0.75

LRFD

  2.00

rn  0.75  77.8 kips/bolt 

ASD

rn 77.8 kips/bolt   2.00  38.9 kips/bolt

 58.4 kips/bolt

The available tearout strength of the flange at the interior bolts is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration. lc  s  d h

 3 in.  , in.  2.06 in. rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2  2.06 in. 0.570 in. 65 ksi   91.6 kips/bolt  = 0.75

LRFD

rn = 0.75  91.6 kips/bolt   68.7 kips/bolt

  2.00

ASD

rn 91.6 kips/bolt   2.00  45.8 kips/bolt

Therefore, bolt shear controls over bearing or tearout at interior bolts. The available tearout strength of the flange at the edge bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration.

lc  lev  0.5  d h   12 in.  0.5 , in.  1.03 in. rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2 1.03 in. w in. 58 ksi   53.8 kips/bolt  = 0.75

LRFD

rn = 0.75  53.8 kips/bolt   40.4 kips/bolt

  2.00

rn 53.8 kips/bolt   2.00  26.9 kips/bolt

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Therefore, bolt shear controls over bearing or tearout at edge bolts. The strength of the bolt group in the flange is determined by summing the strength of the individual fasteners as follows: LRFD Rn   8 bolts  24.3 kips/bolt   194 kips  168 kips

ASD

Rn 

o.k.

  8 bolts 16.2 kips/bolt   130 kips  112 kips o.k.

Tensile strength of the flange plate

The moment arm between flange forces, dm, used for verifying the tensile strength of the flange plate is equal to the depth of the beam plus one plate thickness. This represents the distance between the centerlines of the flange plates at the top and bottom of the beam. From AISC Manual Equation 12-1a or 12-1b, the flange force is: LRFD

ASD

M Puf  u dm

M Paf  a dm





 252 kip-ft 12 in./ft 

18.0 in.  w in.  161 kips

168 kip-ft 12 in./ft 

18.0 in.  w in.  108 kips

From AISC Specification Section J4.1(a), the available tensile yield strength of the flange plate is determined as follows: Ag  bt   7 in. w in.  5.25 in.2

Rn  Fy Ag

(Spec. Eq. J4-1)

  36 ksi   5.25 in.

2



 189 kips

 = 0.90

LRFD

  1.67

Rn  0.90 189 kips   170 kips  161 kips

ASD

Rn 189 kips   1.67  113 kips  108 kips

o.k.

o.k.

From AISC Specification Section J4.1(b), the available tensile rupture strength of the flange plate is determined as follows:

An  b  n  d h  z-in.  t  7 in.   2 bolts , in.  z in.   w in.  3.75 in.2

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Table D3.1, Case 1, applies in this case because the tension load is transmitted directly to the cross-sectional element by fasteners; therefore, U = 1.0. Ae  AnU



2

 3.75 in.

(Spec. Eq. D3-1)

 1.0 

 3.75 in.2 Rn  Fu Ae



  58 ksi  3.75 in.2

(Spec. Eq. J4-2)



 218 kips

 = 0.75

LRFD

  2.00

Rn  0.75  218 kips   164 kips  161 kips

ASD

Rn 218 kips   2.00  109 kips  108 kips

o.k.

o.k.

Flange plate block shear rupture

There are three cases for which block shear rupture of the flange plate must be checked. Case 1, as shown in Figure II.B-1-2(a), involves the tearout of the two blocks outside the two rows of bolt holes in the flange plate; for this case leh = 12 in. and lev = 12 in. Case 2, as shown in Figure II.B-1-2(b), involves the tearout of the block between the two rows of the holes in the flange plate. AISC Manual Tables 9-3a, 9-3b, and 9-3c may be adapted for this calculation by considering the 4 in. width to be comprised of two, 2-in.-wide blocks, where leh = 2 in. and lev = 12 in. Case 1 is more critical than the Case 2 because leh is smaller. Case 3, as shown in Figure II.B-1-2(c), involves a shear failure through one row of bolts and a tensile failure through the two bolts closest to the column. Therefore, Case 1 and Case 3 will be verified. Flange plate block shear rupture—Case 1

The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60Fu Anv  Ubs Fu Ant  0.60Fy Agv  Ubs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the flange plate is determined as follows, using AISC Manual Tables 9-3a, 9-3b and 9-3c, and AISC Specification Equation J4-5, with n = 4, leh = lev = 12 in., and Ubs = 1.0. LRFD Tension rupture component from AISC Manual Table 9-3a:  F A  u nt  43.5 kip/in. t

ASD Tension rupture component from AISC Manual Table 9-3a:





Fu Ant  29.0 kip/in.  t

 Shear yielding component from AISC Manual Table 9-3b:  0.60 Fy Agv  170 kip/in.  t

Shear yielding component from AISC Manual Table 9-3b:





0.60 Fy Agv  113 kip/in. t

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LRFD Shear rupture component from AISC Manual Table 9-3c:

ASD Shear rupture component from AISC Manual Table 9-3c:

0.60 Fu Anv  183 kip/in.   t







The design block shear rupture strength is:

The allowable block shear rupture strength is:





Rn  0.60 Fu Anv  U bs Fu Ant

0.60 Fu Anv  122 kip/in. t

Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant  +   122 kip/in.    2 planes  w in.     1.0  29.0 kip/in. 

 0.60 Fy Agv  U bs Fu Ant 183 kip/in.    2 planes  w in.   1.0 43.5 kip/in.    170 kip/in.    2 planes  w in.     1.0  43.5 kip/in.   340 kips  320 kips

113 kip/in.    2 planes  w in.     1.0  29.0 kip/in.   227 kips  213 kips

  Therefore: Rn  320 kips  161 kips

Therefore: o.k.

Rn  213 kips  108 kips 

o.k.

Flange plate block shear rupture—Case 3 Because AISC Manual Table 9-3a does not include a large enough edge distance, the nominal strength for the limit state of block shear rupture is calculated by directly applying the provisions of AISC Specification Section J4.3.

Rn  0.60Fu Anv  Ubs Fu Ant  0.60Fy Agv  Ubs Fu Ant

Fig. II.B-1-2. Three cases for block shear rupture.

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(Spec. Eq. J4-5)

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where Agv   n  1 s  lev  t   4  1 3 in.  12 in.  w in.  7.88 in.2 Anv  Agv   n  0.5  d h  z in. t

 7.88 in.2 –  4  0.5 , in.  z in. w in.  5.26 in.2

Ant   gage  leh  1.5  d h  z in.  t   4 in.  12 in.  1.5 , in.  z in.   w in.  3.00 in.2 U bs  1.0

and















Rn  0.60  58 ksi  5.26in.2  1.0  58 ksi  3.00 in.2  0.60  36 ksi  7.88 in.2  1.0  58 ksi  3.00 in.2



 357 kips  344 kips

Therefore: Rn  344 kips

From AISC Specification Section J4.3, the available strength for the limit state of block shear rupture on the plate is:   0.75

LRFD

Rn  0.75  344 kips   258 kips  161 kips

  2.00

Rn 344 kips   2.00  172 kips  108 kips

o.k.

ASD

o.k.

Beam flange block shear rupture

The nominal strength for the limit state of block shear rupture is given by AISC Specification Section J4.3.

Rn  0.60Fu Anv  Ubs Fu Ant  0.60Fy Agv  Ubs Fu Ant

(Spec. Eq. J4-5)

The available block shear rupture strength of the beam flange involves the tearout of the two blocks outside the two rows of bolt holes in the flanges. Conservatively use the flange forces that were found for the fastener checks. From AISC Manual Tables 9-3a, 9-3b, and 9-3c, and AISC Specification Equation J4-5, with n = 4, leh = 1w in., lev = 14 in. (reduced 4 in. to account for beam underrun), and Ubs = 1.0:

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LRFD Tension rupture component from AISC Manual Table 9-3a:  F A  u nt  60.9 kip/in. t

ASD Tension rupture component from AISC Manual Table 9-3a:





Fu Ant  40.6 kip/in. t

Shear yielding component from AISC Manual Table 9-3b:  0.60 Fy Agv   231 kip/in. t

Shear yielding component from AISC Manual Table 9-3b:

Shear rupture component from AISC Manual Table 9-3c:

Shear rupture component from AISC Manual Table 9-3c:

0.60 Fu Anv   197 kip/in.  t











0.60 Fy Agv  154 kip/in. t



0.60 Fu Anv  132 kip/in. t

The design block shear rupture strength is:

The allowable block shear rupture strength is:

Rn  0.60 Fu Anv  U bs Fu Ant

Rn 0.60Fu Anv U bs Fu Ant = +    0.60Fy Agv U bs Fu Ant  +   132 kip/in.    2 planes  0.570 in.     1.0  40.6 kip/in.  154 kip/in.    2 planes  0.570 in.     1.0  40.6 kip/in.   197 kips  222 kips

 0.60 Fy Agv  U bs Fu Ant 197 kip/in.    2 planes  0.570 in.      1.0  60.9 kip/in.   231 kip/in.    2 planes  0.570 in.     1.0  60.9 kip/in.   294 kips  333 kips

Therefore:

Therefore:

Rn  294 kips  168 kips

o.k.

Rn  197 kips  112 kips 

o.k.

Fillet weld to supporting column flange The applied load is perpendicular to the weld length (  90); therefore, the directional strength factor is determined from AISC Specification Equation J2-5. This increase factor due to directional strength is incorporated into the weld strength calculation. 1.0  0.50sin1.5   1.0  0.50sin1.5  90   1.50

The required fillet weld size is determined using AISC Manual Equations 8-2a or 8-2b as follows:

LRFD

ASD

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Dmin  

Puf  2 welds 1.50 1.392 kip/in. l 161 kips  2 welds 1.50 1.392 kip/in. 7 in.

 5.51

Dmin  

Paf  2 welds 1.50  0.928 kip/in. l 108 kips  2 welds 1.50  0.928 kip/in. 7 in.

 5.54

Use a a-in. fillet weld on both sides of the flange plate.

Use a a-in. fillet weld on both sides of the flange plate.

Compression Flange Plate and Connection From AISC Specification Section J4.4, the available strength of the flange plate in compression is determined as follows: K = 0.65, from AISC Specification Commentary Table C-A-7.1 L = 3.00 in. (the distance between adjacent bolt holes) r 

I A

 7 in. w in.3 12  7 in. w in.

 0.217 in. Lc KL  r r 0.65  3.00 in.  0.217 in.  8.99

Since Lc/r ≤ 25:

Pn  Fy Ag

(Spec. Eq. J4-6)

  36 ksi  7 in. w in.  189 kips  = 0.90

LRFD

Pn  0.90 189 kips   170 kips  161 kips

o.k.

  1.67

ASD

Pn 189 kips   1.67  113 kips  108 kips o.k.

The compression flange plate will be identical to the tension flange plate; a w-in.7-in. plate with eight bolts in two rows of four bolts on a 4-in. gage and a-in. fillet welds to the supporting column flange. Note: The bolt bearing and shear checks are the same as for the tension flange plate and have found to be adequate in prior calculations. Tension due to load reversal must also be considered in the design of the fillet weld to the supporting column flange. The result is the same as previously calculated for the top flange connection plate. Flange Local Bending of Column

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From AISC Specification Section J10.1, the available strength of the column for the limit state of flange local bending is determined as follows:

0.15b f  0.15 14.6 in.  2.19 in. The length of loading (i.e., plate width) is 7 in., which is greater than 0.15bf. Thus, flange local bending needs to be checked. Assume the concentrated force to be resisted is applied at a distance from the column end greater than 10tf.

10t f  10  0.780 in.  7.80 in. Rn  6.25Fyf t f 2

(Spec. Eq. J10-1)

 6.25  50 ksi  0.780 in.

2

 190 kips

 = 0.90

LRFD

  1.67

Rn  0.90 190 kips   171 kips  161 kips

ASD

Rn 190 kips   1.67  114 kips  108 kips

o.k.

o.k.

Web Local Yielding of Column Assume the concentrated force to be resisted is applied at a distance from the column end that is greater than the depth of the column. The available strength of the column for the limit state of web local yielding is determined from AISC Manual Table 9-4 and AISC Manual Equation 9-47a or 9-47b, with lb = t = w in. LRFD

ASD

R1  83.7 kips

R1   55.8 kips

R2  24.3 kip/in.

R2   16.2 kip/in.

Rn  2  R1   lb  R2 

Rn  2  R1    lb  R2     2  55.8 kips    w in.16.2 kip/in.

 2  83.7 kips    w in. 24.3 kip/in.  186 kips  161 kips o.k.

 124 kips  108 kips o.k.

Web Local Crippling Assume the concentrated force to be resisted is applied at a distance from the column end that is greater than or equal to one-half of the column depth. The available strength of the column for the limit state of web local crippling is determined from AISC Manual Table 9-4 and AISC Manual Equation 9-50a or 9-50b, with lb = t = w in.

LRFD

ASD

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R3  108 kips

R3   71.8 kips

R4  11.2 kip/in.

R4   7.44 kip/in.

Rn  2 R3  lb  R4    2 108 kips   w in.11.2 kip/in. 

Rn  2  R3   lb  R4      2  71.8 kips   w in. 7.44 kip/in. 

 233 kips > 161 kips o.k.

 155 kips  108 kips o.k.

Note: Web compression buckling (AISC Specification Section J10.5) must be checked if another beam is framed into the opposite side of the column at this location. Web panel zone shear (AISC Specification Section J10.6) should also be checked for this column. For further information, see AISC Design Guide 13 Stiffening of Wide-Flange Columns at Moment Connections: Wind and Seismic Applications (Carter, 1999).

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EXAMPLE II.B-2 WELDED FLANGE-PLATED FR MOMENT CONNECTION (BEAM-TO-COLUMN FLANGE) Given: Verify a welded flange-plated FR moment connection between an ASTM A992 W1850 beam and an ASTM A992 W1499 column flange, as shown in Figure II.B-2-1, to transfer the following beam end reactions: Vertical shear: VD = 7 kips VL = 21 kips Strong-axis moment: MD = 42 kip-ft ML = 126 kip-ft Use 70-ksi electrodes. The flange plates are ASTM A36 material. Assume the top flange of the beam is in the tension condition due to moment.

Fig. II.B-2-1. Connection geometry for Example II.B-2.

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Solution: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi Plates ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1850 d = 18.0 in. bf = 7.50 in. tf = 0.570 in. tw = 0.355 in. Zx = 101 in.3 Column W1499 d = 14.2 in. bf = 14.6 in. tf = 0.780 in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2  7 kips   1.6  21 kips 

ASD

Ra  7 kips  21 kips  28.0 kips

 42.0 kips M u  1.2  42 kip-ft   1.6 126 kip-ft 

M a  42 kip-ft  126 kip-ft  168 kip-ft

 252 kip-ft

Single-Plate Web Connection The single-plate web connection is verified in Example II.B-1. Note: By inspection, the available effective fastener strength and shear yielding strengths of the beam web are adequate. The beam web is nearly as thick as the web plate and of a higher strength material. Shear rupture and block shear rupture are not limit states for the beam web. Tension Flange Plate and Connection Tensile yielding of the flange plate The top flange plate is specified as a PL1 in. 6 in. 0 ft 102 in. The top beam flange width is bf = 7.50 in. This provides a shelf dimension of w-in. on both sides of the plate for welding.

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The moment arm between flange plate forces, dm, used for verifying the plate strength is equal to the depth of the beam plus one-half the thickness of each of the flange plates. This represents the distance between the centerlines of the flange plates at the top and bottom of the beam.

d m  18.0 in. 

w in. 1 in.  2 2

 18.9 in. From AISC Manual Equation 12-1a or 12-1b, the flange force is: LRFD

ASD

M Puf  u dm

M Paf  a dm





 252 kip-ft 12 in./ft 

18.9 in.  160 kips

168 kip-ft 12 in./ft 

18.9 in.  107 kips

From AISC Specification Section J4.1(a), the available tensile yield strength of the flange plate is determined as follows: Rn  Fy Ag

(Spec. Eq. J4-1)

  36 ksi  6 in.1 in.  216 kips

  0.90 

LRFD

  1.67 

Rn  0.90  216 kips   194 kips  160 kips

ASD

Rn 216 kips   1.67  129 kips  107 kips

o.k.

o.k.

Fillet weld strength for top flange plate to beam flange The moment arm between flange forces, dm, used for verifying the fillet weld strength is equal to the depth of the beam. This dimension represents the faying surface between the flange of the beam and the tension plate. From AISC Manual Equation 12-1a or 12-1b, the flange force is: LRFD Puf  

Mu dm

 252 kip-ft 12 in./ft 

18.0 in.  168 kips

ASD M Paf  a dm 

168 kip-ft 12 in./ft 

18.0 in.  112 kips

A c-in. fillet weld is specified (D = 5). The available strength may be calculated using the provisions from AISC Specification Section J2.4(b)(2). The available shear strength of the fillet weld may be calculated using AISC Specification Table J2.5. The length of the longitudinally loaded welds is determined taking into consideration a 4-in. tolerance to account for possible beam underrun and a weld termination equal to the weld size.

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l  102 in.  1 in. (setback)  4 in. (underrun)  c in. (weld termination)  8.94 in.

 2  D  Rnwl  0.60 FEXX     l  2   16   2  5   0.60  70 ksi       8.94 in. 2 welds   2   16   166 kips  2  D  Rnwt  0.60 FEXX     l  2   16   2  5   0.60  70 ksi       6 in.  2   16   55.7 kips The combined strength of the fillet weld group may be taken as the larger of the following: Rn  Rnwl  Rnwt  166 kips  55.7 kips  222 kips

(Spec. Eq. J2-6a)

Rn  0.85Rnwl  1.5Rnwt

(Spec. Eq. J2-6b)

 0.85 166 kips   1.5  55.7 kips   225 kips Therefore: Rn = 225 kips   0.75 

LRFD

  2.00 

Rn  0.75  225 kips   169 kips  168 kips

ASD

Rn 225 kips   2.00  113 kips  112 kips o.k.

o.k.

Connecting elements rupture strength at top flange welds At the top flange connection, the minimum base metal thickness to match the shear rupture strength of the weld is determined as follows:

tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  5 

65 ksi  0.238 in. < 0.570 in. beam flange o.k.

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tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  5 

58 ksi  0.266 in.  1.00 in. top flange plate o.k. Fillet weld at top flange plate to column flange The applied load is perpendicular to the weld length (  90), therefore the directional strength factor is determined from AISC Specification Equation J2-5. This increase factor due to directional strength is incorporated into the weld strength calculation. 1.0  0.50sin1.5   1.0  0.50sin1.5  90   1.50

The available strength of fillet welds is determined using AISC Manual Equation 8-2a or 8-2b, as follows:

Dmin

LRFD Puf   2 welds 1.50 1.392 kip/in. l



160 kips  2 welds 1.50 1.392 kip/in. 6 in.

 6.39

ASD Paf Dmin   2 welds 1.50  0.928 kip/in. l



107 kips  2 welds 1.50  0.928 kip/in. 6 in.

 6.41

Use a v-in. fillet weld on both sides of the plate.

Use a v-in. fillet weld on both sides of the plate.

Compression Flange Plate and Connection Flange plate compressive strength The bottom flange plate is specified as a PLw8w1'-22". The bottom flange width is bf = 7.50 in. This provides a shelf dimension of s-in. on both sides of the plate for welding. Assume an underrun dimension of 4-in. and an additional 2-in. to the start of the weld. K = 0.65 from AISC Specification Commentary Table C-A-7.1 L = 1.75 in. r 

I A

8w in. w in.3 12 8w in. w in.

 0.217 in. Lc KL  r r 0.65 1.75 in.  0.217 in.  5.24  25

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Since Lc/r ≤ 25:

Pn  Fy Ag

(Spec. Eq. J4-6)

  36 ksi  8w in. w in.  236 kips LRFD

  0.90

  1.67 

Pn  0.90  236 kips   212 kips  160 kips

ASD

Pn 236 kips   1.67  141 kips  107 kips o.k.

o.k.

Fillet weld strength for bottom flange plate to beam flange The required weld length is determined using AISC Manual Equation 8-2a or 8-2b, as follows: LRFD lmin  

ASD

Pfu 2 welds 1.392 kip/in. D  

lmin 

168 kips  2 welds 1.392 kip/in. 5 



 12.1 in.

Pfa 2 welds 0.928 kip/in. D   112 kips  2 welds  0.928 kip/in. 5 

 12.1 in.

Use 122-in.-long c-in. fillet welds.

Use 122-in.-long c-in. fillet welds.

Beam bottom flange rupture strength at welds Anv   2 welds  t f l   2 welds  0.570 in.122 in.  14.3 in.3 Rn  0.60 Fu Anv



 0.60  65 ksi  14.3 in.

2

(Spec. Eq. J4-4)



 558 kips

  0.75 

LRFD

  2.00

Rn  0.75  558 kips   419 kips  168 kips

ASD

Rn 558 kips   2.00  279 kips  112 kips o.k.

o.k.

Fillet weld at bottom flange plate to column flange The applied load is perpendicular to the weld length (  90) therefore the directional strength factor is determined from AISC Specification Equation J2-5. This increase factor due to directional strength is incorporated into the weld strength calculation.

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1.0  0.50sin1.5   1.0  0.50sin1.5  90   1.50

The available strength of fillet welds is determined using AISC Manual Equation 8-2a or 8-2b as follows: LRFD

Dmin  

ASD

Puf

 2 welds 1.50 1.392 kip/in. l 160 kips  2 welds 1.50 1.392 kip/in. 8w in.

Dmin  

Paf

 2 welds 1.50  0.928 kip/in. l 107 kips  2 welds 1.50  0.928 kip/in. 8w in.

 4.38 sixteenths

 4.39 sixteenths

Use c-in. fillet welds.

Use c-in. fillet welds.

See Example II.B-1 for checks of the column under concentrated forces. For further information, see AISC Design Guide 13 Stiffening of Wide-Flange Columns at Moment Connections: Wind and Seismic Applications. (Carter, 1999). Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.B-3 DIRECTLY WELDED FLANGE FR MOMENT CONNECTION (BEAM-TO-COLUMN FLANGE) Given: Verify a directly welded flange FR moment connection between an ASTM A992 W1850 beam and an ASTM A992 W1499 column flange, as shown in Figure II.B-3-1, to transfer the following beam end reactions: Vertical shear: VD = 7 kips VL = 21 kips Strong-axis moment: MD = 42 kip-ft ML = 126 kip-ft Use 70-ksi electrodes. Check the column for stiffening requirements.

Fig. II.B-3-1. Connection geometry for Example II.B-3. Solution: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam and column ASTM A992 Fy = 50 ksi Fu = 65 ksi

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Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From ASCE/SEI 7, Chapter 2 , the required strength is: LRFD Ru  1.2  7 kips   1.6  21 kips 

ASD

Ra  7 kips  21 kips  28.0 kips

 42.0 kips M u  1.2  42 kip-ft   1.6 126 kip-ft 

M a  42 kip-ft  126 kip-ft  168 kip-ft

 252 kip-ft

The single-plate web connection is verified in Example II.B-1. Note: By inspection, the available effective fastener strength and shear yielding strengths of the beam web are adequate. The beam web is nearly as thick as the web plate, and of a higher strength material. Shear rupture and block shear rupture are not limit states for the beam web. Weld of Beam Flange to Column A complete-joint-penetration groove weld will transfer the entire flange force in tension and compression. It is assumed that the beam is adequate for the applied moment and will carry the tension and compression forces through the flanges. See Example II.B-1 for checks of the column under concentrated forces. For further information, see AISC Design Guide 13 Stiffening of Wide-Flange Columns at Moment Connections: Wind and Seismic Applications. (Carter, 1999). Conclusion The connection is found to be adequate as given for the applied loads.

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CHAPTER IIB DESIGN EXAMPLE REFERENCES Carter, C.J. (1999), Stiffening of Wide-Flange Columns at Moment Connections: Wind and Seismic Applications, Design Guide 13, AISC, Chicago, IL.

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Chapter IIC Bracing and Truss Connections The design of bracing and truss connections is covered in Part 13 of the AISC Steel Construction Manual.

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EXAMPLE II.C-1 TRUSS SUPPORT CONNECTION Given: The truss end connection shown in Figure II.C-1-1 is designed for the required forces shown in Figure II.C-1-2. Verify the following: a. The connection requirements between the gusset and the column b. The required gusset size and the weld requirements connecting the diagonal to the gusset Use 70-ksi electrodes. The top chord and column are ASTM A992 material. The diagonal member, gusset plate and clip angles are ASTM A36 material.

Fig. II.C-1-1. Truss support connection.

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Fig. II.C-1-2. Required forces in members. Solution: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Column and top chord ASTM A992 Fy = 50 ksi Fu = 65 ksi Diagonal, gusset plate and clip angles ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-1, 1-7, 1-8 and 1-15, the geometric properties are as follows: Top chord WT838.5

d = 8.26 in. tw = 0.455 in. y = 1.63 in. Column W1250

d = 12.2 in. tf = 0.640 in. bf = 8.08 in. tw = 0.370 in. Diagonal brace 2L432a t = a in. A = 5.36 in.2 x = 0.947 in. for single angle

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Clip angles 2L44s t = s in. From Figure II.C-1-2 the required strengths are: LRFD

ASD

Brace axial load:

Brace axial load:

Ru  168 kips

Ra  112 kips

Truss end reaction:

Truss end reaction:

Ru  106 kips

Ra  70.4 kips

Top chord axial load:

Top chord axial load:

Ru  131 kips

Ra  87.2 kips

Weld Connecting the Diagonal to the Gusset Plate Note: AISC Specification Section J1.7, requiring that the center of gravity of the weld group coincide with the center of gravity of the member, does not apply to end connections of statically loaded single-angle, double-angle and similar members. From AISC Specification Table J2.4, the minimum fillet weld size for a-in. angles attached to a 2-in.-thick gusset plate is: wmin  x in.

For 4-in. fillet welds (D = 4), the required weld length is determined from AISC Manual Equations 8-2a or 8-2b, as follows: LRFD lreq  

Ru

 4 welds 1.392 kip/in. D  168 kips  4 welds 1.392 kip/in. 4 

 7.54 in.

ASD lreq  

Ra

 4 welds  0.928 kip/in. D  112 kips  4 welds  0.928 kip/in. 4 

 7.54 in.

Use an 8-in.-long 4-in. fillet weld at the heel and toe of each angle. Gusset Shear Rupture at Brace Welds The minimum plate thickness to match the shear rupture strength of the welds is determined as follows:

tmin  

6.19 D Fu

(Manual Eq. 9-3)

6.19  4 

58 ksi  0.427

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Try a 2-in.-thick gusset plate. Tensile Strength of the Brace From AISC Specification Section D2, the available tensile yielding strength of the brace is determined as follows: (Spec. Eq. D2-1)

Pn  Fy Ag





  36 ksi  5.36 in.2  193 kips

LRFD

ASD

t  0.90 

t  1.67 

 t Pn  0.90 193 kips 

Pn 193 kips  t 1.67  116 kips  112 kips o.k.

 174 kips  168 kips o.k.

From AISC Specification Section D2, the available tensile rupture strength of the brace is determined as follows: An  Ag  5.36 in.2 The shear lag factor, U, is determined from AISC Specification Table D3.1, Case 4: U 

3l 2

 x 1   l 3l  w  2

2

3  8 in.

2

3  8 in.   4 in. 2

 0.947 in.  1   8 in.  

2

 0.814

Ae  AnU



2

 5.36 in.

(Spec. Eq. D3-1)

  0.814

 4.36 in.2 Pn  Fu Ae



  58 ksi  4.36 in.2

(Spec. Eq. D2-2)



 253 kips

LRFD t  0.75 

t Pn  0.75  253 kips   190 kips  168 kips o.k.

ASD t  2.00   Pn 253 kips  t 2.00  127 kips  112 kips o.k.

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Use a 2-in.-thick gusset plate. With the brace-to-gusset welds determined, a gusset plate layout as shown in Figure II.C-1-1 can be made. Strength of the Bolted Connection—Angles From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the individual strengths of the individual fasteners, which may be taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10. The number of d-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) required for shear only is determined as follows: LRFD From AISC Manual Table 7-1, the available bolt shear strength is:

ASD From AISC Manual Table 7-1, the available bolt shear strength is:

rn  24.3 kips/bolt

rn  16.2 kips/bolt 

nmin  

Ru

nmin 

 2 bolts/row  rn 106 kips  2 bolts/row  24.3 kips/bolt 

 2.18 rows



Ra 2 bolts/row   rn  70.4 kips

 2 bolts/row 16.2 kips/bolt 

 2.17 rows

Use 2L44s clip angles with five pairs of bolts. Note the number of rows of bolts is increased to “square off” the gusset plate. The available bearing strength of the angles per bolt is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration:

rn  2.4dtFu

(Spec. Eq. J3-6a)

 2.4  d in. s in. 58 ksi   76.1 kips/bolt   0.75

LRFD

rn  0.75  76.1 kips/bolt   57.1 kips/bolt

  2.00

ASD

rn 76.1 kips/bolt     38.1 kips/bolt

The available tearout strength of the angles at edge bolts is determined from AISC Specification Section J3.10, with dh = , in. for d-in.-diameter bolts from AISC Specification Table J3.3, assuming deformation at service load is a design consideration:

lc  le  0.5dh  12 in.  0.5 , in.  1.03 in.

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rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2 1.03 in. s in. 58 ksi   44.8 kips/bolt

  0.75

LRFD

rn  0  44.8 kips/bolt   33.6 kips/bolt

  2.00

ASD

rn 44.8 kips/bolt     22.4 kips/bolt

Therefore, bolt shear controls over bolt bearing or tearout at the edge bolts. The available tearout strength of the angles at interior bolts is determined from AISC Specification Section J3.10, assuming deformation at service load is a design consideration:

lc  s  d h  3 in.  , in.  2.06 in.

rn  1.2lc tFu

(Spec. Eq. J3-6c)

 1.2  2.06 in. s in. 58 ksi   89.6 kips/bolt

  0.75

LRFD

rn  0  89.6 kips/bolt   67.2 kips/bolt

  2.00

ASD

rn 89.6 kips/bolt     44.8 kips/bolt

Therefore, bolt shear controls over bolt bearing or tearout at the interior bolts. Because bolt shear controls for all the bolts, the connection is acceptable based on previous calculations. Bolt Shear and Tension Interaction—Bolts Connecting Clip Angles to Column The eccentric moment about the work point (w.p.) at the faying surface (face of column flange) is determined using an eccentricity equal to half of the column depth. d 2 12.2 in.  2  6.10 in.

e

The eccentricity normal to the plane of the faying surface is accounted for using the Case II approach in AISC Manual Part 7 for eccentrically loaded bolt groups.

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n  4 bolts (number of bolts above the neutral axis) d m  9.00 in. (moment arm between resultant tension force and resultant compressive force) The maximum tensile force per bolt is determined using AISC Manual Equations 7-14a or 7-14b, as follows: LRFD

ASD

Pe rut  u ndm 

Pe rat  a nd m

106 kips  6.10 in.  4 bolts  9.00 in.



 18.0 kips/bolt

 70.4 kips  6.10 in.  4 bolts  9.00 in.

 11.9 kips/bolt

The required shear stress per bolt is determined as follows:

Ab  0.601 in.2 (from AISC Manual Table 7-1) n  10 bolts LRFD

ASD

R f rv  u nAb 

R f rv  a nAb 106 kips

10 bolts   0.601 in.

2





 17.6 ksi

70.4 kips

10 bolts   0.601 in.2 

 11.7 ksi

The nominal tensile strength modified to include the effects of shear stress is determined from AISC Specification Section J3.7 as follows. From AISC Specification Table J3.2:

Fnt  90 ksi Fnv  54 ksi LRFD

  0.75

Fnt f rv  Fnt (Spec. Eq. J3-3a) Fnv 90 ksi  1.3  90 ksi   17.6 ksi   90 ksi 0.75  54 ksi 

Fnt  1.3Fnt 

 77.9 ksi  90 ksi

Fnt f rv  Fnt (Spec. Eq. J3-3b) Fnv 2.00  90 ksi   1.3  90 ksi   11.7 ksi   90 ksi 54 ksi  78.0 ksi  90 ksi

Fnt  1.3Fnt 

Therefore:

Therefore:

Fnt  77.9 ksi

Fnt  78.0 ksi

Bc  Fnt Ab



 0.75  77.9 ksi  0.601 in.2

ASD

  2.00



(from Spec. Eq. J3-2)

 35.1 kips/bolt  18.0 kips/bolt

o.k.

Fnt Ab (from Spec. Eq. J3-2)  78.0 ksi  0.601 in.2 2.00  23.4 kips/bolt  11.9 kips/bolt o.k.

Bc 



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Prying Action on Clip Angles From AISC Manual Part 9, the available tensile strength of the bolts in the outstanding angle legs taking prying action into account is determined as follows: a

b f  gage

2 8.08 in.  42 in.  2  1.79 in.

Note: a is calculated based on the column flange width in this case because it is less than the double angle width. b

gage  t p  t

2 42 in.  2 in.  s in.  2  1.69 in.

Note: 14 in. entering and tightening clearance from AISC Manual Table 7-15 is accommodated and the column fillet toe is cleared. d   d   a    a  b   1.25b  b  2   2   d in. d in.  1.79 in.   1.25 1.69 in.  2 2  2.23 in.  2.55 in. o.k. d   b   b  b  2    1.69 in. 

(Manual Eq. 9-23)

(Manual Eq. 9-18) d in. 2

 1.25 in. b a 1.25 in.  2.23 in.  0.561



(Manual Eq. 9-22)

l n 15 in.  5  3.00 in.

p

Check ps 3.00 in.  3.00 in. o.k.

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d p , in.  1 3.00 in.  0.688

  1

(Manual Eq. 9-20)

The angle thickness required to develop the available strength of the bolt with no prying action is determined as follows: LRFD

  0.90

Bc  35.1 kips/bolt (calculated previously)

tc 

4 Bc b pFu

(Manual Eq. 9-26a)

4  35.1 kips/bolt 1.25 in.



ASD

  1.67

Bc  23.4 kips/bolt (calculated previously)

tc  

0.90  3.00 in. 58 ksi 

4 Bc b pFu

(Manual Eq. 9-26b)

1.67  4  23.4 kips/bolt 1.25 in.

 3.00 in. 58 ksi 

 1.06 in.

 1.06 in.

  tc  2  1    1  1     t    1.06 in. 2  1     1 0.688 1  0.561  s in.    1.75

' 

(Manual Eq. 9-28)

Because    1, the angles have insufficient strength to develop the bolt strength, therefore: 2

t  Q    1     tc  2

 s in.    1  0.688   1.06 in.   0.587 The available tensile strength per bolt, taking prying action into account, is determined using AISC Manual Equation 9-27, as follows: LRFD rn  Bc Q   35.1 kips/bolt  0.587   20.6 kips/bolt  18.0 kips/bolt

o.k.

ASD rn  Bc Q    23.4 kips/bolt  0.587   13.7 kips/bolt  11.9 kips/bolt

o.k.

Shear Strength of Clip Angles From AISC Specification Section J4.2(a), the available shear yielding strength of the angles is determined as follows:

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Agv   2 angles  lt   2 angles 15 in. s in.  18.8 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  18.8 in.2



 406 kips

LRFD

  1.00

  1.50

Rn  1.00  406 kips 

ASD

Rn 406 kips   1.50  271 kips  70.4 kips o.k.

 406 kips  106 kips o.k.

From AISC Specification Section J4.2, the available shear rupture strength of the angles is determined using the net area determined in accordance with AISC Specification Section B4.3b. Anv   2 angles  l  n  d h  z in.  t   2 angles  15 in.  5 , in.  z in.   s in.  12.5 in.2

Rn  0.60 Fu Anv



 0.60  58 ksi  12.5 in.

2

(Spec. Eq. J4-4)



 435 kips   0.75

LRFD

  2.00

Rn  0.75  435 kips 

ASD

Rn 435 kips   2.00  218 kips  70.4 kips o.k.

 326 kips  106 kips o.k. Block Shear Rupture of Clip Angles

The available strength for the limit state of block shear rupture of the angles is determined as follows.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant where Agv   2 angles  l  lev  t   2 angles 15 in.  12 in. s in.  16.9 in.2

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Anv  Agv   2 angles  n  0.5  d h  z in. t  16.9 in.2   2 angles  5  0.5 , in.  z in. s in.  11.3 in.2 Ant   2 angles  leh  0.5  d h  z in.  t   2 angles   2 in.  0.5 , in.  z in.   s in.  1.88 in.2 U bs  1.0

and















Rn  0.60  58 ksi  11.3 in.2  1.0  58 ksi  1.88 in.2  0.60  36 ksi  16.9 in.2  1.0  58 ksi  1.88 in.2



 502 kips  474 kips

Therefore: Rn  474 kips

  0.75

LRFD

  2.00

ASD

Rn 474 kips   2.00  237 kips  70.4 kips o.k.

Rn  0.75  474 kips   356 kips  106 kips o.k. Prying Action on Column Flange

Using the same procedure as shown previously for the clip angles, the available tensile strength of the bolts, taking prying action into account, is: LRFD Tc  18.7 kips  18.0 kips o.k.

ASD Tc  12.4 kips  11.9 kips o.k.

Strength of the Bolted Connection—Column Flange By inspection, the applicable limit states will control for the angles; therefore, the column flange is acceptable. Clip Angle-to-Gusset Plate Connection With a top chord slope of 2 in 12, the horizontal welds are unequal length as shown in Figure II.C-1-3. The average horizontal length is used in the following calculations.

l  15 in. 3a in.  2w in. 2  3.06

kl 

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kl l 3.06 in.  15 in.  0.204

k

 kl 2 l  2  kl   3.06 in.2  15 in.  2  3.06 in.

xl 

 0.443 in.

al  xl  6.10 in.  4.00 in.  10.1 in. 10.1 in.  xl l 10.1 in.  0.443 in.  15 in.  0.644

a

By interpolating AISC Manual Table 8-8 with Angle = 0:

C  1.50

Fig. II.C-1-3. Weld group geometry.

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From AISC Manual Table 8-8, the minimum required weld size is determined as follows: LRFD

  0.75 

Dmin  

  2.00 

Ru  2 welds  CC1l

Dmin 

106 kips 2 welds 0.75   1.50 1.0 15 in.

 3.14



ASD

Ra 2 welds   CC1l 2.00  70.4 kips 

2 1.50 1.0 15 in.

 3.13

Use 4-in. fillet welds.

Use 4-in. fillet welds.

From AISC Specification Table J2.4, the minimum weld size for s-in. clip angles attached to a 2-in.-thick gusset plate is: wmin  x in.  4 in.

o.k.

Note: Using the average of the horizontal weld lengths provides a reasonable solution when the horizontal welds are close in length. A conservative solution can be determined by using the smaller of the horizontal weld lengths as effective for both horizontal welds. For this example, use kl = 2w in., C = 1.43, and Dmin = 3.29 sixteenths. Tensile Yielding of Gusset Plate on the Whitmore Section The gusset plate thickness should match or slightly exceed that of the chord stem. This requirement is satisfied by the 2-in. plate previously selected. From AISC Manual Figure 9-1, the width of the Whitmore section is:

lw  4.00 in.  2  8.00 in. tan 30  13.2 in. From AISC Specification Section J4.1(a), the available tensile yielding strength of the gusset plate is determined as follows: Ag  lwt  13.2 in.2 in.  6.60 in.2 Rn  Fy Ag

(Spec. Eq. J4-1)



  36 ksi  6.60 in.2



 238 kips

  0.90

LRFD

Rn  0.90  238 kips   214 kips  168 kips o.k.

  1.67 

ASD

Rn 238 kips   1.67  143 kips  112 kips o.k.

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Gusset Plate-to-Tee Stem Weld The interface forces are: LRFD Horizontal shear between gusset and WT:

ASD Horizontal shear between gusset and WT:

H ub  131 kips   4 bolts 18.0 kips/bolt 

H ab  87.2 kips   4 bolts 11.9 kips/bolt 

 59.0 kips

 39.6 kips

Vertical tension between gusset and WT:

Vertical tension between gusset and WT:

 4 bolts  Vub  106 kips     10 bolts   42.4 kips

 4 bolts  Vab   70.4 kips     10 bolts   28.2 kips

Compression between WT and column:

Compression between WT and column:

Cub   4 bolts 18.0 kips/bolt 

Cab   4 bolts 11.9 kips/bolt 

 72.0 kips

 47.6 kips

Summing moments about the face of the column at the workline of the top chord:

Summing moments about the face of the column at the workline of the top chord:

M ub  Cub  22 in.  1.50 in.

M ab  Cab  22 in.  1.50 in.

 H ub  d  y 

 H ab  d  y 

 gusset width   Vub   setback  2     72.0 kips  22 in.  1.50 in.   59.0 kips  8.26 in.  1.63 in.  15.0 in.    42.4 kips    2 in.  2    340 kip-in.

 gusset width   Vab   setback  2     47.6 kips  22 in.  1.50 in.   39.6 kips  8.26 in.  1.63 in.  15.0 in.    28.2 kips    2 in.  2    227 kip-in.

A CJP weld should be used along the interface between the gusset plate and the tee stem. The weld should be ground smooth under the clip angles. The gusset plate width depends upon the diagonal connection. From a scaled layout, the gusset plate must be 1 ft 3 in. wide. The gusset plate depth depends upon the connection angles. From a scaled layout, the gusset plate must extend 12 in. below the tee stem. Use a PL212 in.1 ft 3 in. Conclusion The connection is found to be adequate as given for the applied loads.

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EXAMPLE II.C-2 TRUSS SUPPORT CONNECTION Given:

Verify the truss support connections, as shown in Figure II.C-2-1, at the following joints: A. Joint L1 B. Joint U1 Use 70-ksi electrodes, ASTM A36 plate, ASTM A992 bottom and top chords, and ASTM A36 double angles.

Fig. II.C-2-1. Connection geometry for Example II.C-2. Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Top and bottom chord ASTM A992 Fy = 50 ksi Fu = 65 ksi

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Web member, diagonal members and plate ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-7, 1-8 and 1-15, the geometric properties are as follows: Top Chord WT838.5

tw = 0.455 in. d = 8.26 in. Bottom Chord WT828.5 tw = 0.430 in. d = 8.22 in. Diagonal U0L1

2L432a

A = 5.36 in.2 x  0.947 in. (for single angle)

Web U1L1

2L323c

A = 3.90 in.2

Diagonal U1L2

2L3222c

A = 3.58 in.2 x  0.632 in. (for single angle)

As shown in Figure II.C-2-1, the required forces are: LRFD

ASD

Web U1L1 load:

Web U1L1 load:

Pu  104 kips

Pa  69.2 kips

Diagonal U0L1 load:

Diagonal U0L1 load:

Tu  +165 kips

Ta  +110 kips

Diagonal U1L2 load:

Diagonal U1L2 load:

Tu  +114 kips

Ta  +76 kips

Solution A:

Shear Yielding of Bottom Chord Stem From AISC Specification Section J4.2(a), the available shear yielding strength of the bottom chord at Section A-A (see Figure II.C-2-1) is determined as follows:

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Agv  dtw   8.22 in. 0.430 in.  3.53 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  3.53 in.2



 106 kips

  1.00 

LRFD

  1.50 

Rn  1.00 106 kips 

ASD

Rn 106 kips   1.50  70.7 kips  69.2 kips

 106 kips  104 kips o.k.

o.k.

Welds for Member U1L1 Note: AISC Specification Section J1.7 requiring that the center of gravity of the weld group coincide with the center of gravity of the member does not apply to end connections of statically loaded single angle, double angle and similar members. From AISC Specification Table J2.4, the minimum weld size for a c-in.-thick angle is: wmin  x in.

From AISC Specification Section J2.2b(b)(2), the maximum weld size is: wmax  t  z in.  c  z in.  4 in.

Try a x in. fillet weld. The minimum weld length is determined using AISC Manual Equation 8-2a or 8-2b:

lmin

LRFD Ru   2 sides  2 welds 1.392 kip/in. D 

104 kips  2 sides  2 welds 1.392 kip/in. 3

 6.23 in.

lmin

ASD Ra   2 sides  2 welds  0.928 kip/in. D 

69.2 kips  2 sides  2 welds  0.928 kip/in. 3

 6.21 in.

Use a 62-in.-long weld at the heel and toe of the angles.

Use a 62-in.-long weld at the heel and toe of the angles.

Shear Rupture Strength of Angles at Welds The minimum angle thickness to match the required shear rupture strength of the welds is determined as follows:

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tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  3

58 ksi  0.160 in.  c in. o.k. Shear Rupture Strength of Tee-Stem at Welds The minimum tee-stem thickness to match the required shear rupture strength of the welds is determined as follows:

tmin  

6.19 D Fu

(Manual Eq. 9-3)

6.19  3

65 ksi  0.286 in.  0.430 in. o.k. Note, both the top and bottom chords are acceptable for x-in. fillet welds. Welds for Member U0L1 From AISC Specification Table J2.4, the minimum weld size for a a-in.-thick angle is: wmin  x in.

From AISC Specification Section J2.2b(b)(2), the maximum weld size is: wmax  t  z in.  a  z in.  c in.

Try a x in. fillet weld. The minimum weld length is determined using AISC Manual Equation 8-2a or 8-2b:

lmin

LRFD Ru   2 sides  2 welds 1.392 kip/in. D 

165 kips  2 sides  2 welds 1.392 kip/in. 3

 9.88 in.

ASD lmin  

Ra

 2 sides  2 welds  0.928 kip/in. D 110 kips  2 sides  2 welds  0.928 kip/in. 3

 9.88 in.

Use a 10-in.-long weld at the heel and toe of the angles.

Use a 10-in.-long weld at the heel and toe of the angles.

Note: A plate will be welded to the stem of the WT to provide room for the connection. Based on the preceding calculations for the minimum angle and stem thicknesses, by inspection the angles, stems, and stem plate extension have adequate strength. Tensile Strength of Diagonal U0L1 From AISC Specification Section D2, the available tensile yielding strength of the angles is determined as follows:

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(Spec. Eq. D2-1)

Pn  Fy Ag





  36 ksi  5.36 in.

2

 193 kips

LRFD

ASD

t  0.90 

t  1.67 

t Pn  0.90 193 kips 

Pn 193 kips  t 1.67  116 kips  110 kips

 174 kips  165 kips o.k.

o.k.

From AISC Specification Section D2, the available tensile rupture strength of the angles is determined as follows. The shear lag factor, U, is determined using AISC Specification Table D3.1, Case 4. U 

3l 2

 x 1   l 3l  w2  2

3 10 in.

2

3 10 in.   4 in. 2

2

 0.947 in.  1   10 in.  

 0.859 Pn  Fu Ae



  58 ksi  5.36 in.

2

(Spec. Eq. D2-2)

  0.859

 267 kips

LRFD

ASD

t  0.75 

 t  2.00 

t Pn  0.75  267 kips 

Pn 267 kips  t 2.00  134 kips  110 kips

 200 kips  165 kips o.k.

o.k.

Block Shear Rupture of Bottom Chord The available strength for the limit state of block shear rupture of the chord is determined as follows.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant where Agv  Anv   2 lines  lt w   2 lines 10 in. 0.430 in.  8.60 in.2

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(Spec. Eq. J4-5)

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Ant   angle leg  t   4 in. 0.430 in.  1.72 in.2 U bs  1.0

Note, because ASTM A36 is used for the stem extension plate, Fy = 36 ksi and Fu = 58 ksi are used for the shear components of AISC Specification Equation J4-5.















Rn  0.60  58 ksi  8.60 in.2  1.0  65 ksi  1.72 in.2  0.60  36 ksi  8.60 in.2  1.0  65 ksi  1.72 in.2



 411 kips  298 kips

Therefore: Rn  298 kips

LRFD

  0.75

Rn  0.75  298 kips   224 kips  165 kips o.k.

  2.00

ASD

Rn 298 kips   2.00  149 kips  110 kips o.k.

Solution B:

Shear Yielding of Top Chord Stem From AISC Specification Section J4.2(a), the available shear yielding strength of the top chord at Section B-B (see Figure II.C-2-1) is determined as follows: Agv  dtw   8.26 in. 0.455 in.  3.76 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  50 ksi  3.76 in.

2



 113 kips

  1.00 

LRFD

Rn  1.00 113 kips   113 kips  74.0 kips o.k.

  1.50 

ASD

Rn 113 kips   1.50  75.3 kips  49.2 kips o.k.

Welds for Member U1L1 As calculated previously in Solution A, use 62-in.-long x-in. fillet welds at the heel and toe of both angles.

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Welds for Member U1L2 As determined in previous calculations, the minimum and maximum weld sizes for a c-in.-thick angle are:

wmin  x in. wmax  4 in. Try a 4 in. fillet weld. To avoid having to use a stem extension plate unequal length welds are provided at the heel and toe of the angle. The minimum weld length for each angle is determined using AISC Manual Equation 8-2a or 8-2b:

lmin

LRFD Ru   2 sides 1.392 kip/in. D 

lmin

114 kips

ASD Ra   2 sides  0.928 kip/in. D 

 2 sides 1.392 kip/in. 4 

76 kips

 2 sides  0.928 kip/in. 4 

 10.2 in.

 10.2 in.

Try 72 in. of 4-in. fillet weld at the heel and 4 in. of 4-in. fillet weld at the toe of each angle. l  72 in.  4 in. =11.5 in.  10.2 in.

o.k.

Shear Rupture Strength of Angles at Welds The minimum angle thickness to match the required shear rupture strength of the welds is determined as follows:

tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  4 

58 ksi  0.213 in.  c in. o.k. Shear Rupture Strength of Tee-Stem at Welds The minimum tee-stem thickness to match the required shear rupture strength of the welds is determined as follows:

tmin  

6.19 D Fu

(Manual Eq. 9-3)

6.19  4 

65 ksi  0.381 in.  0.455 in. o.k. Tensile Strength of Diagonal U1L2 From AISC Specification Section J4.1(a), the available tensile yielding strength of the angles are determined as follows:

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(Spec. Eq. J4-1)

Rn  Fy Ag



  36 ksi  3.58 in.2



 129 kips

LRFD

  0.90 

  1.67 

Rn  0.90 129 kips 

ASD

Rn 129 kips   1.67  77.2 kips  76 kips o.k.

 116 kips  114 kips o.k.

From AISC Specification Section J4.1(b), the available tensile rupture strength of the angles are determined as follows. The shear lag factor, U, is determined using AISC Specification Table D3.1, Case 4. l1  l2 2 72 in.  4 in.  2  5.75 in.

l 

U 

3l 2

 x 1   l 3l  w  2

2

3  5.75 in.

2

3  5.75 in.   32 in. 2

2

 0.632 in.  1   5.75 in.  

 0.792 Rn  Fu Ae



  58 ksi  3.58 in.

2

(Spec. Eq. J4-2)

  0.792

 164 kips

  0.75 

LRFD

Rn  0.75 164 kips   123 kips  114 kips o.k.

  2.00 

ASD

Rn 164 kips   2.00  82.0 kips  76 kips o.k.

Conclusion Joints L1 and U1 are found to be adequate as given for the applied loads.

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EXAMPLE II.C-3 HEAVY WIDE-FLANGE COMPRESSION CONNECTION (FLANGES ON THE OUTSIDE) Given:

The truss shown in Figure II.C-3-1 has been designed with ASTM A992 W14 shapes with flanges to the outside of the truss. Beams framing into the top chord and lateral bracing are not shown but can be assumed to be adequate. Based on multiple load cases, the critical dead and live load forces for this connection are shown in Figure II.C-3-2. A typical top chord connection is shown in Figure II.C-3-1, Detail A. Design this typical connection using 1-in.diameter Group A slip-critical bolts in standard holes with threads not excluded from the shear plane (thread condition N) with Class A faying surfaces and ASTM A36 gusset plates.

Fig II.C-3-1. Truss layout for Example II.C-3.

Fig. II.C-3-2. Forces at Detail A.

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Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: W-shapes ASTM A992 Fy = 50 ksi Fu = 65 ksi Gusset plates ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Top chord W14109 d = 14.3 in. bf = 14.6 in. tf = 0.860 in. Web members W1461 d = 13.9 in. bf = 10.0 in. tf = 0.645 in. From AISC Specification Table J3.3, for 1-in.-diameter bolts with standard holes: d h  18 in.

From ASCE/SEI 7, Chapter 2, the required strengths are determined as follows and summarized in Figure II.C-3-2. LRFD

ASD

Left top chord:

Left top chord:

Pu  1.2  262 kips   1.6  262 kips 

Pa  262 kips  262 kips  524 kips

 734 kips Right top chord:

Right top chord:

Pu  1.2  345 kips   1.6  345 kips 

Pa  345 kips  345 kips  690 kips

 966 kips Vertical Web:

Vertical Web:

Pu  1.2 102 kips   1.6 102 kips 

Pa  102 kips  102 kips  204 kips

 286 kips

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LRFD

ASD

Diagonal Web:

Diagonal Web:

Pu  1.2 113 kips   1.6 113 kips 

Pa  113 kips  113 kips  226 kips

 316 kips

Note: In checking equilibrium of vertical forces, Fy  0, due to the external (loading) forces not included. Refer to Figure II.C-3-2 for the magnitude of external load forces. In most truss designs, member forces only are provided and force equilibrium of the internal truss forces will not sum to zero. Bolt Slip Resistance Strength From AISC Specification Section J3.8(a), the available slip resistance for the limit state of slip for standard size holes is determined as follows:   0.30 for Class A surface Du  1.13 h f  1.0, no filler is provided Tb  51 kips, from AISC Specification Table J3.1, Group A ns  1, number of slip planes

  rn  Du h f Tb ns



(Spec. Eq. J3-4)

  0.30 1.131.0  51 kips 1  17.3 kips/bolt   1.00 

LRFD

 rn  1.00 17.3 kips/bolt   17.3 kips/bolt

ASD

  1.50   rn 17.3 kips/bolt   1.50  11.5 kips/bolt



Note: Standard holes are used in both plies for this example. Other hole sizes may be used and should be considered based on the preferences of the fabricator or erector on a case-by-case basis.

(a) LRFD

(b) ASD

Fig. II.C-3-2. Required forces at Detail A.

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Diagonal Connection The required number of bolts is determined as follows: LRFD

ASD

Pu  316 kips

Pa  226 kips

Pa rn 226 kips  11.5 kips/bolt  19.7 bolts

Pu rn 316 kips  17.3 kips/bolt  18.3 bolts

nreq 

nreq 

For two lines of bolts on both sides, the required number of rows is:

For two lines of bolts on both sides, the required number of rows is:

18.3 bolts  4.58  2 sides  2 lines 

19.7 bolts  4.93  2 sides  2 lines 

Therefore, use five rows at min. 3-in. spacing.

Therefore, use five rows at min. 3-in. spacing.

Whitmore section in gusset plate The width of the Whitmore section, lw, is determined as shown in AISC Manual Figure 9-1. lw  gage  2l tan 30  52 in.  2 12 in. tan 30   19.4 in.

Try a a-in.-thick plate. Ag   2 plates  lwt   2 plates 19.4 in. a in.  14.6 in.2

From AISC Specification Section J4.1(a), the available tensile yielding strength of the gusset plate is determined as follows: Rn  Fy Ag

(Spec. Eq. J4-1)



  36 ksi  14.6 in.

2



 526 kips

  0.90

LRFD

Rn  0.90  526 kips   473 kips  316 kips o.k.

  1.67

ASD

Rn 526 kips   1.67  315 kips  226 kips o.k.

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Block shear rupture of gusset plate The available strength for the limit state of block shear rupture of the gusset plates is determined as follows.

Rn  0.60 Fu Anv  U bs Fu Ant  0.60 Fy Agv  U bs Fu Ant

(Spec. Eq. J4-5)

where Agv   2 plates  2 lines  lev   n  1 s  t   2 plates  2 lines   2 in.   5  1 3 in.   a in.  21.0 in.2

Anv  Agv   2 plates  2 lines  5  0.5  d h  z in. t  21.0 in.2   2 plates  2 lines  5  0.5  18 in.  z in. a in.  13.0 in.2 Ant   2 plates   gage   d h  z in.  t   2 plates  52 in.  18 in.  z in.   a in.  3.23 in.2 U bs  1.0

and















Rn  0.60  58 ksi  13.0 in.2  1.0  58 ksi  3.23 in.2  0.60  36 ksi  21.0 in.2  1.0  58 ksi  3.23 in.2



 640 kips  641 kips

Therefore: Rn  640 kips

  0.75

LRFD

  2.00

ASD

Rn 640 kips   2.00  320 kips  226 kips o.k.

Rn  0.75  640 kips   480 kips  316 kips o.k. Block shear rupture of diagonal flange

By inspection, block shear rupture on the diagonal flange will not control. Strength of bolted connection—gusset plate Slip-critical connections must also be designed for the limit states of bearing-type connections. From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the individual strengths of the individual fasteners, which may be taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10.

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From AISC Manual Table 7-1, the available shear strength per bolt for 1-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) is: LRFD

ASD rn  21.2 kips/bolt 

rn  31.8 kips/bolt

The available bearing and tearout strength of the gusset plate at the edge bolts is determined using AISC Manual Table 7-5, using le = 2 in. LRFD

ASD rn   50.0 kip/in. a in.   18.8 kips/bolt

rn   75.0 kip/in. a in.  28.1 kips/bolt

Therefore, the bearing or tearout strength controls over bolt shear at the edge bolts. The available bearing and tearout strength of the gusset plate at the other bolts is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

ASD rn   65.3 kip/in. a in.   24.5 kips/bolt

rn   97.9 kip/in. a in.  36.7 kips/bolt

Therefore, bolt shear controls over bearing or tearout at the other bolts. The strength of the bolt group in the gusset plate is determined by summing the strength of the individual fasteners as follows: LRFD 1 bolt  28.1 kips/bolt 

     4 bolts  31.8 kips/bolt    621 kips  316 kips o.k.

Rn   2 sides  2 lines  

ASD  1 bolt 18.8 kips/bolt   Rn   2 sides  2 lines        4 bolts  21.2 kips/bolt    414 kips  226 kips o.k.

Strength of bolted connection—diagonal flange By inspection the strength of the bolted connection at the gusset plate will control. Horizontal Connection The required strength of the gusset plate to horizontal member is determined as follows: LRFD

Pu  966 kips  734 kips  232 kips

ASD

Pa  690 kips  524 kips  166 kips

Using the bolt slip resistance strength determined previously, the required number of rows of bolts is determined as follows:

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LRFD nreq

ASD

P  u rn 232 kips  17.3 kips/bolt  13.4 bolts

nreq

Pu  rn 166 kips  11.5 kips/bolt  14.4 bolts

For two lines of bolts on both sides the required number of rows is:

For two lines of bolts on both sides the required number of rows is:

13.4 bolts  3.35  2 sides  2 lines 

14.4 bolts  3.60  2 sides  2 lines 

For members not subject to corrosion the maximum bolt spacing is determined using AISC Specification Section J3.5(a):

24t  24  a in.  9.00 in. Due to the geometry of the gusset plate, the use of 4 rows of bolts in the horizontal connection will exceed the maximum bolt spacing; instead use 5 rows of bolts in two lines. Shear strength of the gusset plate From AISC Specification Section J4.2(a), the available shear yielding strength of the gusset plates is determined as follows: Agv   2 plates  lt   2 plates  32.0 in. a in.  24.0 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  24.0 in.2



 518 kips

  1.00

LRFD

  1.50

Rn  1.00  518 kips 

ASD

Rn 518 kips   1.50  345 kips  166 kips o.k.

 518 kips  232 kips o.k.

From AISC Specification Section J4.2(b), the available shear rupture strength of gusset plates is determined as follows: Anv   2 plates  l  n  d h  z in.  t   2 plates  32.0 in.  5 18 in.  z in.   a in.  19.5 in.2

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Rn  0.60 Fu Anv



 0.60  58 ksi  19.5 in.

2

(Spec. Eq. J4-4)



 679 kips

LRFD

  0.75

  2.00

Rn  0.75  679 kips 

ASD

Rn 679 kips   1.50  453 kips  166 kips o.k.

 509 kips  232 kips o.k. Strength of bolted connection

By comparison to the preceding calculations for the diagonal connection, bolt bearing or tearout does not control. Vertical Connection Using the bolt slip resistance strength determined previously, the required number of bolts is determined as follows: LRFD

ASD

Pu  286 kips

Pu  204 kips

Pu rn 204 kips  11.5 kips/bolt  17.7 bolts

Pu rn 286 kips  17.3 kips/bolt  16.5 bolts

nreq 

nreq 

For two lines of bolts on both sides, the required number of rows is:

For two lines of bolts on both sides, the required number of rows is:

16.5 bolts  4.12  2 sides  2 lines 

17.7 bolts  4.43  2 sides  2 lines 

Therefore, use 5 rows at min. 3-in. spacing.

Therefore, use 5 rows at min. 3-in. spacing.

Shear strength of the gusset plate From AISC Specification Section J4.2(a), the available shear yielding strength of gusset plates is determined as follows: Agv   2 plates  lt   2 plates  31w in. a in.  23.8 in.2 Rn  0.60 Fy Agv

(Spec. Eq. J4-3)



 0.60  36 ksi  23.8 in.2



 514 kips

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LRFD

  1.00

  1.50

Rn  1.00  514 kips 

ASD

Rn 514 kips   1.50  343 kips  204 kips o.k.

 514 kips  286 kips o.k.

From AISC Specification Section J4.2(b), the available shear rupture strength of gusset plates is determined as follows: Anv   2 plates  l  n  d h  z in.  t   2 plates  31w in.  7 18 in.  z in.   a in.  17.6 in.2 Rn  0.60 Fu Anv



 0.60  58 ksi  17.6 in.2

(Spec. Eq. J4-4)



 612 kips

  0.75

LRFD

  2.00

Rn  0.75  612 kips   459 kips  286 kips

ASD

Rn 612 kips   2.00  306 kips  204 kips

o.k.

o.k.

Strength of bolted connection By comparison to the preceding calculations for the diagonal connection, bolt bearing does not control. Note that because of the difference in depths between the top chord and the vertical and diagonal members, x-in. loose shims are required on each side of the shallower members. The final connection design is shown in Figure II.C-3-4.

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Fig. II.C-3-4. Connection layout for Example II.C-3.

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Chapter IID Miscellaneous Connections This section contains design examples on connections in the AISC Steel Construction Manual that are not covered in other sections of the AISC Design Examples.

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EXAMPLE II.D-1 WT HANGER CONNECTION Given: Design an ASTM A992 WT hanger connection between an ASTM A36 2L33c tension member and an ASTM A992 W2494 beam to support the following loads: PD = 13.5 kips PL = 40 kips Use 70-ksi electrodes. Solution: From AISC Manual Table 2-4, the material properties are as follows: Beam and WT hanger ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles ASTM A36 Fy = 36 ksi Fu = 58 ksi From AISC Manual Tables 1-1, 1-7 and 1-15, the geometric properties are as follows: Beam W2494

d = 24.3 in. tw = 0.515 in. bf = 9.07 in. tf = 0.875 in. Angles 2L33c A = 3.56 in.2 x = 0.860 in. (for single angle) From AISC Specification Table J3.3, the hole diameter for w-in.-diameter bolts with standard holes is:

d h  m in. From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Pu  1.2 13.5 kips   1.6  40 kips 

 80.2 kips

ASD

Pa  13.5 kips  40 kips  53.5 kips

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Weld Design Note: AISC Specification Section J1.7 requiring that the center of gravity of the weld group coincide with the center of gravity of the member does not apply to end connections of statically loaded single-angle, double-angle and similar members. From AISC Specification Table J2.4, the minimum weld size for a c-in.-thick angle is: wmin  x in.

From AISC Specification Section J2.2b(b)(2), the maximum weld size is: wmax  t  z in.  c  z in.  4 in.

Try 4-in. fillet welds. The minimum weld length is determined using AISC Manual Equations 8-2a or 8-2b, as follows:

lmin

LRFD Ru   2 sides  2 welds 1.392 kip/in. D 

80.2 kips

 2 sides  2 welds 1.392 kip/in. 4 

 3.60 in.

lmin

ASD Ra   2 sides  2 welds  0.928 kip/in. D 

53.5 kips

 2 sides  2 welds  0.928 kip/in. 4 

 3.60 in.

Use a 4-in.-long weld at the heel and toe of the angles.

Use a 4-in.-long weld at the heel and toe of the angles.

Tensile Strength of Angles From AISC Specification Section D2, the available tensile yielding strength of the angles is determined as follows: Pn  Fy Ag

(Spec. Eq. D2-1)



  36 ksi  3.56 in.2



 128 kips

LRFD

t  0.90   t Pn  0.90 128 kips   115 kips  80.2 kips o.k.

 t  1.67 

ASD

Pn 128 kips  t 1.67  76.6 kips  53.5 kips o.k.

From AISC Specification Section D2, the available tensile rupture strength of the brace is determined as follows: An  Ag

 3.56 in.2 The shear lag factor, U, is determined from AISC Specification Table D3.1, Case 4:

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U 

3l 2

 x 1   l 3l  w  2

2

3  4 in.

2

3  4 in.   3 in. 2

 0.860 in.  1   4 in.  

2

 0.661

Ae  AnU



2

 3.56 in.

(Spec. Eq. D3-1)

  0.661

 2.35 in.2 Pn  Fu Ae



  58 ksi  2.35 in.2

(Spec. Eq. D2-2)



 136 kips

LRFD

ASD  t  2.00   Pn 136 kips  t 2.00  68.0 kips  53.5 kips o.k.

t  0.75  t Pn  0.75 136 kips   102 kips  80.2 kips o.k.

Preliminary WT Selection Using Beam Gage Try four w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N), with a 4-in. gage. LRFD

ASD

Tr  rut

Tr  rat

P  u n 80.2 kips  4 bolts  20.1 kips/bolt

Pa n 53.5 kips  4 bolts  13.4 kips/bolt 

From AISC Manual Table 7-2:

From AISC Manual Table 7-2:

Bc  rn  29.8 kips/bolt  20.1 kips/bolt o.k.

rn   19.9 kips/bolt  13.4 kips/bolt

Bc 

Determine tributary length per pair of bolts, p, using AISC Manual Figure 9-4.

42 in. 8.00 in.  42 in.  2 2  4.00 in.

p

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Check: ps 4.00 in.  42 in.

o.k.

Verify that the tributary length on each side of the bolt conforms to dimensional limits assuming a 2-in. tee stem thickness: b

 4.00 in.  2 in. 2

 1.75 in.

42 in.  1.75b 2 2.25 in.  3.06 in. o.k. 8.00 in.  42 in.  1.75b 2 1.75 in.  3.06 in. o.k. A preliminary hanger connection is determined using AISC Manual Table 15-2b.

2 Rut  

 rows  Bc

LRFD

p

 2  20.1 kips/bolt 

4.00 in.  10.1 kip/in.

2 Rat  

 rows  Bc

ASD

p

 2 bolts 13.4 kips/bolt 

4.00 in.  6.70 kip/in.

From AISC Manual Table 15-2b, with an assumed b = (4.00 in. – 2 in.)/2 = 1.75 in., the flange thickness, t = tf, of the WT hanger should be approximately s in. The minimum depth WT that can be used is equal to the sum of the weld length plus the weld size plus the kdimension for the selected section. From AISC Manual Table 1-8 with an assumed b = 1.75 in., tf  s in., and dmin = 4 in. + 4 in. + k  6 in., appropriate selections include: WT625 WT726.5 WT825 WT927.5

Try a WT625. From AISC Manual Table 1-8, the geometric properties are as follows: bf = 8.08 in. tf = 0.640 in. tw = 0.370 in.

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Prying Action From AISC Manual Part 9, the available tensile strength of the bolts taking prying action into account is determined as follows. The beam flange is thicker than the WT flange; therefore, prying in the tee flange will control over prying in the beam flange. a

b f  gage

2 8.08 in.  4 in.  2  2.04 in.

gage  t w 2 4 in.  0.370 in.  2  1.82 in.

b

b  b 

db 2

(Manual Eq. 9-18)

 w in.   1.82 in.     2   1.45 in.

d  a   a  b 2 

db      1.25b   2    w in. w in.  2.04 in.   1.25 1.82 in.  2 2  2.42 in.  2.65 in.

(Manual Eq. 9-23)

b a 1.45 in.  2.42 in.  0.599



(Manual Eq. 9-22)

From AISC Manual Equation 9-21: LRFD 1B     c  1   Tr  

1  29.8 kips/bolt   1  0.599  20.1 kips/bolt 

 0.806

ASD 1B     c  1   Tr  

1  19.9 kips/bolt   1  0.599  13.4 kips/bolt 

 0.810

d   dh  m in.

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d p m in.  1 4.00 in.  0.797

  1

(Manual Eq. 9-20)

Because   1.0 : LRFD

ASD

1        1.0  1  

1        1.0   1  

1  0.806     1.0 0.797  1  0.806   5.21  1.0

1  0.810     1.0 0.797  1  0.810   5.35  1.0





Therefore,   1.0.

Therefore,   1.0.

  0.90

  1.67

tmin 



4Tu b pFu 1    

(Manual Eq. 9-19a)

4  20.1 kips/bolt 1.45 in.

0.90  4.00 in. 65 ksi  1   0.797 1.0  

tmin 



 4Ta b pFu 1    

(Manual Eq. 9-19b)

1.67  4 13.4 kips/bolt 1.45 in.

 4.00 in. 65 ksi  1   0.797 1.0 

 0.527 in.  t f  0.640 in. o.k.

 0.527 in.  t f  0.640 in. o.k.

Note: As an alternative to the preceding calculations, the designer can use a simplified procedure to select a WT hanger with a flange thick enough to eliminate prying action. Assuming b = 1.45 in., the required thickness to eliminate prying action is determined from AISC Manual Equation 9-17a or 9-17b, as follows: LRFD

  0.90 tnp  

ASD

  1.67

4Tu b pFu

tnp 

4  20.1 kips/bolt 1.45 in.

=

0.90  4.00 in. 65 ksi 

 4Ta b pFu 1.67  4 13.4 kips/bolt 1.45 in.

 4.00 in. 65 ksi 

= 0.707 in.

 0.706 in.

The WT625 that was selected does not have a sufficient flange thickness to reduce the effect of prying action to an insignificant amount. In this case, the simplified approach requires a WT section with a thicker flange. Tensile Yielding of the WT Stem on the Whitmore Section As shown in AISC Manual Figure 9-1, the Whitmore section defines the effective width of the WT stem. Note that the Whitmore section cannot exceed the actual 8 in. width of the WT.

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



lw  3.00 in.  2  4.00 in. tan 30  8.00 in. = 7.62 in.  8.00 in.

Therefore: lw  7.62 in.

From AISC Specification Section J4.1(a), the available tensile yielding strength of the WT stem is determined as follows: Ag  lwtw   7.62 in. 0.370 in.  2.82 in.2

Rn  Fy Ag

(Spec. Eq. J4-1)



  50 ksi  2.82 in.

2



 141 kips

  0.90

LRFD

  1.67

ASD

Rn 141 kips   1.67  84.4 kips  53.5 kips o.k.

Rn  0.90 141 kips   127 kips  80.2 kips o.k.

Shear Rupture of the WT Stem Base Metal

From AISC Specification Section J4.2(b), the available shear rupture strength of the WT stem at the welds is determined as follows:

Rn   2 welds  2 planes  0.60 Fu lwtw   2 welds  2 planes  0.60  65 ksi  4 in. 0.370 in.

(from Spec. Eq. J4-4)

 231 kips

  0.75

LRFD

  2.00

ASD

Rn 231 kips   2.00  116 kips  53.5 kips o.k.

Rn  0.75  231 kips   173 kips  80.2 kips o.k.

Block Shear Rupture of the WT Stem The available strength for the limit state of block shear rupture of the stem is determined as follows.

Rn  0.60Fu Anv  Ubs Fu Ant  0.60Fy Agv  U bs Fu Ant where

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Agv  Anv   2 lines  ltw   2 lines  4 in. 0.370 in.  2.96 in.2 Ant   leg  t w   3 in. 0.370 in.  1.11 in.2 U bs  1.0

and















Rn  0.60  65 ksi  2.96 in.2  1.0  65 ksi  1.11 in.2  0.60  50 ksi  2.96 in.2  1.0  65 ksi  1.11 in.2  188 kips  161 kips

Therefore: Rn  161 kips

  0.75

LRFD

  2.00

Rn  0.75 161 kips 

ASD

Rn 161 kips   2.00  80.5 kips  53.5 kips o.k.

 121 kips  80.2 kips o.k. The final connection design is shown in Figure II.D-1-1.

Fig. II.D-1-1. Final hanger design for Example II.D-1

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EXAMPLE II.D-2

BEAM BEARING PLATE

Given:

An ASTM A992 W1850 beam supported by a 10-in.-thick concrete wall, as shown in Figure II.D-2-1, has the following end reactions: RD = 15 kips RL = 45 kips Verify the following: A. If a bearing plate is required when the beam is supported by the full wall thickness (lb = h = 10 in) B. The bearing plate required if lb = h = 10 in. (the full wall thickness) C. The bearing plate required if lb = 62 in. and the bearing plate is centered on the thickness of the wall The concrete has fc = 3 ksi and the bearing plate is ASTM A36 material.

Fig. II.D-2-1. Connection geometry for Example II.D-2. Solution:

From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Bearing plate ASTM A36 Fy = 36 ksi Fu = 58 ksi Concrete wall fc = 3 ksi From AISC Manual Table 1-1, the geometric properties are as follows:

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Beam W1850 d = 18.0 in. tw = 0.355 in. bf = 7.50 in. tf = 0.570 in. kdes = 0.972 in. k1 = m in. From ASCE/SEI, Chapter 2, the required strength is: LRFD Ru  1.2 15 kips   1.6  45 kips 

ASD

Ra  15 kips  45 kips  60.0 kips

 90.0 kips Solution A:

Required Bearing Length The required bearing length for the limit state of web local yielding is determined using AISC Manual Table 9-4 and AISC Manual Equation 9-46a or 9-46b, as follows: LRFD

ASD

R1   28.8 kips R2   11.8 kip/in.

R1  43.1 kips R2  17.8 kip/in.

Ru  R1  kdes R2 90.0 kips  43.1 kips   0.972 in. 17.8 kip/in.  2.63 in.  0.972 in.

lb min 

lb min 

Ra  R1   kdes R2 

60.0 kips  28.8 kips  0.972 in. 11.8 kip/in.  2.64 in.  0.972 in. 

Therefore:

Therefore:

lb min  2.63 in.  10.0 in. o.k.

lb min  2.64 in.  10.0 in. o.k.

The required bearing length for the limit state of web local crippling is determined using AISC Manual Table 9-4.

lb 10.0 in.  d 18.0 in.  0.556 Because

lb > 0.2, use AISC Manual Table 9-4 and AISC Manual Equation 9-49a or 9-49b, as follows: d

LRFD

R5  52.0 kips R6  6.30 kip/in.

ASD

R5   34.7 kips R6   4.20 kip/in.

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lb min

LRFD Ru  R5   kdes R6 90.0 kips  52.0 kips   0.972 in. 6.30 kip/in.  6.03 in.  0.972 in.

ASD lb min

R  R5   a  kdes R6  60.0 kips  34.7 kips  0.972 in. 4.20 kip/in.  6.02 in.  0.972 in. 

Therefore:

Therefore:

lb min  6.03 in.  10.0 in. o.k.

lb min  6.02 in.  10.0 in. o.k.

Verify

lb > 0.2: d

Verify

lb 6.03 in.  d 18.0 in.  0.335  0.2

lb > 0.2: d

lb 6.02 in.  d 18.0 in.  0.334  0.2

o.k.

o.k.

The bearing strength of the concrete is determined from AISC Specification Section J8. Note that AISC Specification Equation J8-1 is used because A2 is not larger than A1 in this case. A1  b f lb   7.50 in.10.0 in.  75.0 in.2

Pp  0.85 f cA1

(Spec. Eq. J8-1)



 0.85  3 ksi  75.0 in.2



 191 kips

LRFD

ASD

c  0.65

c  2.31

c Pp  0.65 191 kips 

Pp

 124 kips  90.0 kips o.k.

191 kips c 2.31  82.7 kips  60.0 kips o.k. 

Beam Flange Thickness Using the cantilever length from AISC Manual Part 14, determine the minimum beam flange thickness required if no bearing plate is provided. The beam flanges along the length, n, are assumed to be fixed end cantilevers with a minimum thickness determined using the limit state of flexural yielding. n

bf

 kdes 2 7.50 in.   0.972 in. 2  2.78 in.

(from Manual Eq. 14-1)

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LRFD The bearing pressure is determined as follows: fp =

Ru A1

fp =

The required flexural strength of the flange is: Mu 

ASD The bearing pressure is determined as follows:

f p n2

Ra A1

The required flexural strength of the flange is: Ma 

f p n2

2 R n2  a 2 A1

2 R n2  u 2 A1

The available flexural strength of the flange is:

The available flexural strength of the flange is:

  0.90

  1.67

M n  Fy Z

M n Fy Z    Fy  t f 2     4

tf 2  Fy   4 

   

For Rn  Ru and solving for tf, the minimum flange thickness is determined as follows: tf

min





2 Ru n 2 A 1 Fy



0.90 75.0 in.2

For Rn   Ra and solving for tf, the minimum flange thickness is determined as follows: tf

2  90.0 kips  2.78 in.

min



2



  50 ksi 

 0.642 in.  t f  0.570 in.

n.g.

 Therefore, a bearing plate is required.

   

 2 Ra n 2 A 1 Fy 1.67  2  60.0 kips  2.78 in.

2

 75.0 in.   50 ksi  2

 0.643 in.  t f  0.570 in.

n.g.

 Therefore, a bearing plate is required.

Note: The designer may assume a bearing width narrower than the beam flange to justify a thinner flange. In this case, the bearing width is constrained by the lower bound concrete bearing strength and the upper bound 0.570-in. flange thickness. 5.43 in. ≤ bearing width ≤ 6.56 in.

Solution B: Bearing Length From Solution A, with lb = 10 in., the web local yielding and web local crippling strengths for the beam are adequate. Bearing Plate Design The required bearing plate width is determined using AISC Specification Equation J8-1 as follows:

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LRFD

ASD

c  2.31

c  0.65

Ru c 0.85 fc  90.0 kips  0.65  0.85 3 ksi 

A1 req 

A1 req 



A1 req

Breq 

lb 2

A1 req lb

54.4 in.2 10.0 in.  5.44 in.

54.3 in. 10.0 in.  5.43 in. 

 60.0 kips  2.31 0.85  3 ksi 

 54.4 in.2

 54.3 in.2 Breq 

Ra c 0.85 f c 



Use B = 8 in. (selected as the least whole-inch dimension that exceeds bf).

Use B = 8 in. (selected as the least whole-inch dimension that exceeds bf).

From AISC Manual Part 14, the bearing plate cantilever dimension is determined as follows: B  kdes 2 8 in.   0.972 in. 2  3.03 in.

n

(Manual Eq. 14-1)

The required thickness of the base plate is determined using the available flexural strength equation previously derived for the required beam flange thickness. LRFD tmin  

ASD

2 Ru n 2 Fy Blb 2  90.0 kips  3.03 in.

tmin  2

0.90  36 ksi  8 in.10 in.

 0.798 in.

Use PLd in.10 in.0 ft 8 in.



 2 Ra n 2 Fy Blb 1.67  2  60.0 kips  3.03 in.

2

 36 ksi  8 in.10 in.

 0.799 in.

Use PLd in.10 in.0 ft 8 in.

Note: The calculations for tmin are conservative. Taking the strength of the beam flange into consideration results in a thinner required bearing plate or no bearing plate at all.

Solution C: From Solution A, with lb = 62 in., the web local yielding and web local crippling strengths for the beam are adequate. Bearing Plate Design

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Try B = 8 in. A1  Blb   8 in. 62 in.  52.0 in.2

AISC Specification Section J8 requires that the area, A2, be geometrically similar to A1.

N1  62 in.  2 1.75 in.  10.0 in. B1  8 in.  2 1.75 in.  11.5 in. A2  B1 N1  11.5 in.10.0 in.  115 in.2

The bearing strength of the concrete is determined from AISC Specification Section J8. Note that AISC Specification Equation J8-2 is used because A2 is larger than A1 in this case. Pp  0.85 f c  A1 A2 A1  1.7 f c  A1



 0.85  3 ksi  52.0 in.2



(Spec. Eq. J8-2)



115 in.2 52.0 in.2  1.7  3 ksi  52.0 in.2



 197 kips  265 kips

Therefore: Pp  197 kips

LRFD

ASD

c  0.65

c  2.31

c Pp  0.65 197 kips 

Pp 197 kips  2.31 c  85.3 kips  60.0 kips o.k.

 128 kips  90.0 kips o.k.

From AISC Manual Part 14, the bearing plate cantilever dimension is determined as follows: B  kdes 2 8 in.   0.972 in. 2  3.03 in.

n

(Manual Eq. 14-1)

The required thickness of the base plate is determined using the available flexural strength equation previously derived for the required beam flange thickness.

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ASD

LRFD tmin  

2

2 Ru n Fy Blb 2  90.0 kips  3.03 in.

tmin  2

0.90  36 ksi  8 in. 62 in.

 0.990 in.

Use PL1 in.62 in.0 ft 8 in.



 2 Ra n Fy Blb

2

1.67  2  60.0 kips  3.03 in.

2

 36 ksi 8 in. 62 in.

 0.991 in.

Use PL1 in.62 in.0 ft 8 in

Note: The calculations for tmin are conservative. Taking the strength of the beam flange into consideration results in a thinner required bearing plate or no bearing plate at all.

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EXAMPLE II.D-3 SLIP-CRITICAL CONNECTION WITH OVERSIZED HOLES Given: Verify the connection of an ASTM A36 2L33c tension member to an ASTM A36 plate welded to an ASTM A992 beam, as shown in Figure II.D-3-1, for the following loads: PD = 15 kips PL = 45 kips

Fig. II.D-3-1. Connection configuration for Example II.D-3. Solution: From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Hanger and plate ASTM A36 Fy = 36 ksi Fu = 58 ksi

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From AISC Manual Tables 1-1, 1-7 and 1-15, the geometric properties are as follows: Beam W1626 tf = 0.345 in. tw = 0.250 in. kdes = 0.747 in. Hanger 2L33c

A = 3.56 in.2 x  0.860 in. for single angle From AISC Specification Table J3.3, the hole diameter for w-in.-diameter bolts with standard and oversized holes is:

d h  m in. (standard hole) d h  , in. (oversized hole) From ASCE/SEI 7, Chapter 2, the required strength is: LRFD Ru  1.2 15 kips   1.6  45 kips 

ASD

Ra  15 kips  45 kips  60.0 kips

 90.0 kips Bolt Slip Resistance Strength

From AISC Manual Table 7-3, with w-in.-diameter Group A slip-critical bolts with Class A faying surfaces in oversized holes and double shear, the available slip resistance strength is: ASD

LRFD rn  10.8 kips/bolt 

rn  16.1 kips/bolt The required number of bolts is determined as follows:

ASD

LRFD

R n u rn 90.0 kips  16.1 kips/bolt  5.59

Ra n r  n / 

Therefore, use 6 bolts.

Therefore, use 6 bolts.

60.0 kips 10.8 kips/bolt  5.56 

Strength of Bolted Connection—Angles Slip-critical connections must also be designed for the limit states of bearing-type connections. From the Commentary to AISC Specification Section J3.6, the strength of the bolt group is taken as the sum of the individual strengths of the individual fasteners, which may be taken as the lesser of the fastener shear strength per AISC Specification Section J3.6, the bearing strength at the bolt hole per AISC Specification Section J3.10, or the tearout strength at the bolt hole per AISC Specification Section J3.10.

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From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear is: LRFD

ASD

rn  23.9 kips/bolt 

rn  35.8 kips/bolt

The available bearing and tearout strength of the angles using standard holes at the edge bolt is determined using AISC Manual Table 7-5, conservatively using le = 14 in. LRFD

ASD

rn   2 angles  44.0 kip/in. c in.  27.5 kips/bolt

rn 

  2 angles  29.4 kip/in. c in.  18.4 kips/bolt

Therefore, the bearing or tearout strength controls over bolt shear at the edge bolts. The available bearing and tearout strength of the angles using standard holes at the other bolts is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

ASD

rn   2 angles  78.3 kip/in. c in.  48.9 kips/bolt

rn 

  2 angles  52.2 kip/in. c in.  32.6 kips/bolt

Therefore, bolt shear controls over bearing or tearout at the other bolts. The strength of the bolt group in the angles is determined by summing the strength of the individual fasteners as follows: LRFD

Rn  1 bolt  27.5 kips/bolt    5 bolts  35.8 kips/bolt   207 kips  90.0 kips o.k.

ASD

Rn 

 1 bolt 18.4 kips/bolt    5 bolts  23.9 kips/bolt   138 kips  60.0 kips o.k.

Tensile Strength of the Angles

From AISC Specification Section J4.1(a), the available tensile yielding strength of the angles is determined as follows: Pn  Fy Ag

(Spec. Eq. J4-1)



  36 ksi  3.56 in.2



 128 kips

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LRFD

  0.90

  1.67

ASD

Pn 128 kips   1.67  76.6 kips  60.0 kips

Pn  0.90 128 kips   115 kips > 90.0 kips o.k.

o.k.

From AISC Specification Section J4.1(b), the available tensile rupture strength of the angles is determined as follows. The shear lag factor, U, is determined using AISC Specification Table D3.1, Case 2. x l 0.860 in.  1 15.0 in.  0.943

U  1

Ae  AnU

(Spec. Eq. D3-1)

  Ag   2 angles  dh  z in. t  U  3.56 in.2   2 angles m in.  z in. c in.   0.943  2.84 in.2 Pn  Fu Ae



  58 ksi  2.84 in.

2

(Spec. Eq. J4-2)



 165 kips

  0.75

LRFD

  2.00

ASD

Pn 165 kips   2.00  82.5 kips  60.0 kips

Pn  0.75 165 kips   124 kips > 90.0 kips o.k.

o.k.

Block Shear Rupture Strength of the Angles

The available strength for the limit state of block shear rupture of the angles is determined as follows:

Rn  0.60Fu Anv  Ubs Fu Ant  0.60Fy Agv  Ubs Fu Ant where Agv   2 angles  lev   n  1 s  t   2 angles  12 in.   6  1 3 in.   c in.  10.3 in.2

Anv  Agv   2 angles  n  0.5  d h  z in. t  10.3 in.2   2 angles  6  0.5  m in.  z in. c in.  7.29 in.2

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(Spec. Eq. J4-5)

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Ant   2 angles  leh  0.5  d h  z in.  t   2 angles  14 in.  0.5 m in.  z in.   c in.  0.508 in.2 U bs  1.0

and















Rn  0.60  58 ksi  7.29 in.2  1.0  58 ksi  0.508 in.2  0.60  36 ksi  10.3 in.2  1.0  58 ksi  0.508 in.2



 283 kips  252 kips

Therefore: Rn  252 kips   0.75

LRFD

Rn  0.75  252 kips   189 kips  90.0 kips o.k.

  2.00

ASD

Rn 252 kips   2.00  126 kips  60.0 kips o.k.

Strength of Bolted Connection—Plate From AISC Manual Table 7-1, the available shear strength per bolt for w-in.-diameter Group A bolts with threads not excluded from the shear plane (thread condition N) in double shear is: LRFD

ASD

rn  23.9 kips/bolt 

rn  35.8 kips/bolt

The available bearing and tearout strength of the plate using oversized holes at the edge bolt is determined using AISC Manual Table 7-5, conservatively using le = 14 in. LRFD

rn   40.8 kip/in.2 in.  20.4 kips/bolt

ASD

rn 

  27.2 kip/in.2 in.  13.6 kips/bolt

Therefore, the bearing or tearout strength controls over bolt shear at the edge bolts. The available bearing and tearout strength of the plate using oversized holes at the other bolts is determined using AISC Manual Table 7-4 with s = 3 in. LRFD

rn   78.3 kip/in.2 in.  39.2 kips/bolt

ASD

rn 

  52.2 kip/in.2 in.  26.1 kips/bolt

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Therefore, bolt shear controls over bearing or tearout at the other bolts. The strength of the bolt group in the plate is determined by summing the strength of the individual fasteners as follows: LRFD

ASD

Rn

Rn  1 bolt  20.4 kips/bolt 



  5 bolts  35.8 kips/bolt   199 kips  90.0 kips o.k.

 1 bolt 13.6 kips/bolt    5 bolts  23.9 kips/bolt   133 kips  60.0 kips o.k.

Tensile Strength of the Plate From AISC Specification Section J4.1(a), the available tensile yielding strength of the plate is determined as follows. By inspection, the Whitmore section, as defined in AISC Manual Figure 9-1, includes the entire width of the 2-in. plate. Ag  bt   6 in.2 in.  3.00 in.2 Rn  Fy Ag

(Spec. Eq. J4-1)



  36 ksi  3.00 in.2



 108 kips

LRFD

  0.90

  1.67

ASD

Pn 108 kips   1.67  64.7 kips  60.0 kips

Rn  0.90 108 kips   97.2 kips > 90.0 kips o.k.

o.k.

From AISC Specification Section J4.1(b), the available tensile rupture strength of the plate is determined as follows: An  Ag   d h + z in. t  3.00 in.2  , in. + z in.2 in.  2.50 in.2

AISC Specification Table D3.1, Case 1, applies in this case because tension load is transmitted directly to the crosssectional element by fasteners; therefore, U = 1.0.

Ae  AnU



2

 2.50 in.

(Spec. Eq. D3-1)

 1.0

 2.50 in.2

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Rn  Fu Ae



  58 ksi  2.50 in.2

(Spec. Eq. J4-2)



 145 kips LRFD

  0.75

ASD

  2.00

Pn 145 kips   2.00  72.5 kips  60.0 kips

Rn  0.75 145 kips   109 kips > 90.0 kips o.k.

o.k.

Block Shear Rupture Strength of the Plate The available strength for the limit state of block shear rupture of the plate is determined as follows.

Rn  0.60Fu Anv  Ubs Fu Ant  0.60Fy Agv  Ubs Fu Ant

(Spec. Eq. J4-5)

where Agv  lev   n  1 s  t  12 in.   6  1 3 in.  2 in.  8.25 in.2

Anv  Agv   n  0.5  d h  z in. t  8.25 in.2   6  0.5  , in.  z in.2 in.  5.50 in.2 Ant  leh  0.5  d h  z in.  t  3 in.  0.5 , in.  z in.  2 in.  1.25 in.2 U bs  1.0

and















Rn  0.60  58 ksi  5.50 in.2  1.0  58 ksi  1.25 in.2  0.60  36 ksi  8.25 in.2  1.0  58 ksi  1.25 in.2  264 kips  251 kips

Therefore: Rn  251 kips

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LRFD

  0.75

  2.00

Rn  0.75  251 kips 

ASD

Rn 251 kips   2.00  126 kips  60.0 kips o.k.

 188 kips  90.0 kips o.k.

Plate-to-Beam Weld The applied load is perpendicular to the weld length (  90), therefore the directional strength factor is determined from AISC Specification Equation J2-5. This increase factor due to directional strength is incorporated into the weld strength calculation. 1.0  0.50 sin1.5   1.0  0.50 sin1.5  90   1.50

The required fillet weld size is determined using AISC Manual Equation 8-2a or 8-2b, as follows:

Dreq

LRFD Pu   2 welds 1.501.392 kip/in. l



Dreq

90.0 kips 2 welds 1.50   1.392 kip/in. 6 in.

ASD Pa   2 welds 1.50 0.928 kip/in. l



 3.59

60.0 kips 2 welds 1.50    0.928 kip/in. 6 in.

 3.59

Use 4-in. fillet welds on each side of the plate.

Use 4-in. fillet welds on each side of the plate.

From AISC Manual Table J2.4, the minimum fillet weld size is:

wmin  x in.  4 in. o.k. Beam Flange Base Metal Check The minimum flange thickness to match the required shear rupture strength of the welds is determined as follows:

tmin  

3.09 D Fu

(Manual Eq. 9-2)

3.09  3.59 

65 ksi  0.171 in.  0.345 in. o.k. Beam Concentrated Forces Check From AISC Specification Section J10.2, the beam web is checked for the limit state of web local yielding assuming the connection is at a distance from the member end greater than the depth of the member, d. Rn  Fyw tw  5kdes  lb 

(Spec. Eq. J10-2)

  50 ksi  0.250 in. 5  0.747 in. + 6 in.  122 kips

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  1.00

LRFD

Rn  1.00 122 kips   122 kips  90.0 kips o.k.

  1.50

ASD

Rn 122 kips   1.50  81.3 kips  60.0 kips o.k.

Conclusion The connection is found to be adequate as given for the applied loads.

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Part III System Design Examples

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EXAMPLE III-1 DESIGN OF SELECTED MEMBERS AND LATERAL ANALYSIS OF A FOURSTORY BUILDING INTRODUCTION This section illustrates the load determination and selection of representative members that are part of the gravity and lateral frame of a typical four-story building. The design is completed in accordance with the AISC Specification and AISC Manual. Loading criteria are based on ASCE/SEI 7. This section includes:  Analysis and design of a typical steel frame for gravity loads  Analysis and design of a typical steel frame for lateral loads  Examples illustrating three methods for satisfying the stability provisions of AISC Specification Chapter C The building being analyzed in this design example is located in a Midwestern city with moderate wind and seismic loads. The loads are given in the description of the design example. All members are ASTM A992 material. CONVENTIONS The following conventions are used throughout this example: 1.

Beams or columns that have similar, but not necessarily identical, loads are grouped together. This is done because such grouping is generally a more economical practice for design, fabrication and erection.

2.

Certain calculations, such as design loads for snow drift, which might typically be determined using a spreadsheet or structural analysis program, are summarized and then incorporated into the analysis. This simplifying feature allows the design example to illustrate concepts relevant to the member selection process.

3.

Two commonly used deflection calculations, for uniform loads, have been rearranged so that the conventional units in the problem can be directly inserted into the equation for design. They are as follows: Simple beam: 

 





5  w kip/in. L in.



4

384  29,000 ksi  I in.4



 w kip/ft  L ft 4



1,290 I in.4



Beam fixed at both ends: 



 w kip/in. L in.4 384  29,000 ksi   I in.4   w kip/ft  L ft 4



6,440 I in.4



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DESIGN SEQUENCE

The design sequence is presented as follows: 1.

General description of the building including geometry, gravity loads and lateral loads

2.

Roof member design and selection

3.

Floor member design and selection

4.

Column design and selection for gravity loads

5.

Wind load determination

6.

Seismic load determination

7.

Horizontal force distribution to the lateral frames

8.

Preliminary column selection for the moment frames and braced frames

9.

Seismic load application to lateral systems

10. Stability (P-) analysis

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GENERAL DESCRIPTION OF THE BUILDING Geometry

The design example is a four-story building, consisting of seven bays at 30 ft in the east-west (numbered grids) direction and bays of 45 ft, 30 ft and 45 ft in the north-south (lettered grids) direction, as shown in Figure III-1. The floor-to-floor height for the four floors is 13 ft 6 in. and the height from the fourth floor to the roof (at the edge of the building) is 14 ft 6 in. Based on discussions with fabricators, the same column size will be used for the whole height of the building. The plans of these floors and the roof are shown on Sheets S2.1 thru S2.3, found at the end of this Chapter. The exterior of the building is a ribbon window system with brick spandrels supported and back-braced with steel and infilled with metal studs. The spandrel wall extends 2 ft above the elevation of the edge of the roof. The window and spandrel system is shown on design drawing Sheet S4.1. The roof system is 12-in. metal deck on open web steel joists. The open web steel joists are supported on steel beams as shown on Sheet S2.3. The roof slopes to interior drains. The middle three bays have a 6-ft-tall screen wall around them and house the mechanical equipment and the elevator over run. This area has steel beams, in place of open web steel joists, to support the mechanical equipment. The three elevated floors have 3 in. of normal weight concrete over 3-in. composite deck for a total slab thickness of 6 in. The supporting beams are spaced at 10 ft on center. These beams are carried by composite girders in the eastwest direction to the columns. There is a 30 ft by 29 ft opening in the second floor, to create a two-story atrium at the entrance. These floor layouts are shown on Sheets S2.1 and S2.2. The first floor is a slab on grade and the foundation consists of conventional spread footings.

Fig. III-1. Basic building layout.

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The building includes both moment frames and braced frames for lateral resistance. The lateral system in the northsouth direction consists of chevron braces at the end of the building located adjacent to the stairways. In the eastwest direction there are no locations in which chevron braces can be concealed; consequently, the lateral system in the east-west direction is composed of moment frames at the north and south faces of the building. This building is sprinklered and has large open spaces around it, and consequently does not require fireproofing for the floors. Wind Forces

The Basic Wind Speed is 107 miles per hour (3-second gust). Because it is sited in an open, rural area, it will be analyzed as Wind Exposure Category C. Because it is an ordinary office occupancy, the building is Risk Category II. Seismic Forces

The sub-soil has been evaluated and the site class has been determined to be Site Class D. The area has a short period Ss = 0.121g and a one-second period S1 = 0.060g. The Seismic Importance Factor is 1.0, that of an ordinary office occupancy (Risk Category II). Roof and Floor Loads

Roof Loads The ground snow load, pg, is 20 psf. The slope of the roof is 4 in./ft or more at all locations, but not exceeding 2 in./ft; consequently, 5 psf rain-on-snow surcharge is to be considered, but ponding instability design calculations are beyond the scope of this example. This roof can be designed as a fully exposed roof, but, per ASCE/SEI 7, Section 7.3, cannot be designed for less than pf = (I)pg = 20 psf uniform snow load. Snow drift will be applied at the edges of the roof and at the screen wall around the mechanical area. The roof live load for this building is 20 psf, but may be reduced per ASCE/SEI 7, Section 4.8, where applicable.

Floor Loads The basic live load for the floor is 50 psf. An additional partition live load of 20 psf is specified, which exceeds the minimum partition load required by ASCE/SEI 7, Section 4.3.2. Because the locations of partitions and, consequently, corridors are not known, and will be subject to change, the entire floor will be designed for a live load of 80 psf. This live load will be reduced based on type of member and area per the ASCE/SEI 7 provisions for liveload reduction. Wall Loads A wall load of 55 psf will be used for the brick spandrels, supporting steel, and metal stud back-up. A wall load of 15 psf will be used for the ribbon window glazing system.

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ROOF MEMBER DESIGN AND SELECTION

Calculate dead load and snow load. Dead load: Roofing Insulation Deck Beams Joists Misc. Total

= 5 psf = 2 psf = 2 psf = 3 psf = 3 psf = 5 psf = 20 psf

Snow load from ASCE/SEI 7, Sections 7.3 and 7.10: Snow Rain on snow Total

= 20 psf = 5 psf = 25 psf

Note: In this design, the rain and snow load is greater than the roof live load. The deck is 12 in., wide rib, 22 gage, painted roof deck, placed in a pattern of three continuous spans minimum. The typical joist spacing is 6 ft on center. At 6 ft on center, this deck has an allowable total load capacity of 87 psf (from the manufacturer’s catalog). The roof diaphragm and roof loads extend 6 in. past the centerline of grid as shown on Sheet S4.1. From ASCE/SEI 7, Section 7.7, the following drift loads are calculated: Flat roof snow load: pg = 20 psf Density:  = 16.6 lb/ft3 hb = 1.20 ft Summary of Drifts

The snow drift at the penthouse was calculated for the maximum effect, using the east-west wind and an upwind fetch from the parapet to the centerline of the columns at the penthouse. This same drift is conservatively used for wind in the north-south direction. The precise location of the drift will depend upon the details of the penthouse construction, but will not affect the final design in this case. A summary of the drift load is given in Table III-1.

Side parapet End parapet Screen wall

Upwind Roof Length, lu, ft 121 211 60.5

Table III-1 Summary of Drifts Projection Height, ft 2 2 6

Max. Drift Load, psf 13.2 13.2 30.5

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Max. Drift Width, W, ft 6.36 6.36 7.35

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SELECT ROOF JOISTS Layout loads and size joists.

The 45-ft side joist with the heaviest loads is shown in Figure III-2 with end reactions and maximum moment. Note: Joists may be specified using ASD or LRFD but are most commonly specified by ASD as shown here.

Fig III-2. Joist loading and bracing diagram—ASD. Because the load is not uniform, select a 24KCS4 joist from the Steel Joist Institute (SJI) Load Tables and Weight Tables for Steel Joists and Joist Girders (SJI, 2015). This joist has an allowable moment of 92.3 kip-ft, an allowable shear of 8.40 kips, a gross moment of inertia of 453 in.4 and weighs 16.5 plf. The first joist away from the end of the building is loaded with snow drift along the length of the member. Based on analysis, a 24KCS4 joist is also acceptable for this uniform load case. As an alternative to directly specifying the joist sizes on the design document, as done in this example, loading diagrams can be included on the design documents to allow the joist manufacturer to economically design the joists. The typical 30-ft-long joist in the middle bay will have a uniform load of: w   6 ft  20 psf  25 psf   270 plf wS   6 ft  25 psf   150 plf

From the SJI load tables, select an 18K5 joist that weighs approximately 7.7 plf and satisfies both strength and deflection requirements. Note: the first joist away from the screen wall and the first joist away from the end of the building carry snow drift. Based on analysis, an 18K7 joist will be used in these locations.

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SELECT ROOF BEAMS

Calculate loads and select beams in the mechanical area. For the beams in the mechanical area, the mechanical units could weigh as much as 60 psf. Use 40 psf additional dead load, which will account for the mechanical units and the screen wall around the mechanical area. Use 15 psf additional snow load, which will account for any snow drift that could occur in the mechanical area. The beams in the mechanical area are spaced at 6 ft on center. Loading is calculated as follows and shown in Figure III-3.



wD   6 ft  0.020 kip/ft 2  0.040 kip/ft 2



 0.360 kip/ft



wS   6 ft  0.025 kip/ft 2  0.015 kip/ft 2



 0.240 kip/ft

Fig. III-3. Loading and bracing diagram for roof beams in mechanical area. From ASCE/SEI 7, Chapter 2, calculate the required strength of the beams in the mechanical area. LRFD wu  1.2  0.360 kip/ft   1.6  0.240 kip/ft   0.816 kip/ft

 30 ft  Ru   0.816 kip/ft     2   12.2 kips

Mu 

 0.816 kip/ft  30 ft 2 8

 91.8 kip-ft

ASD wa  0.360 kip/ft  0.240 kip/ft  0.600 kip/ft

 30 ft  Ra   0.600 kip/ft     2   9.00 kips

Ma 

 0.600 kip/ft  30 ft 2 8

 67.5 kip-ft

As discussed in AISC Design Guide 3, Serviceability Design Considerations for Steel Buildings (West and Fisher, 2003), limit deflection to L/360 because a plaster ceiling will be used in the lobby area.

 30 ft 12 in./ft  L  360 360  1.00 in.

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Using the equation for deflection derived previously, the required moment of inertia, Ix req, can be determined as follows. Use 40 psf as an estimate of the snow load, including some drifting that could occur in this area, for deflection calculations.

I x req 

 0.240 kip/ft  30 ft 4 1,290 1.00 in.

 151 in.4 From AISC Manual Table 3-3, select a beam size with an adequate moment of inertia. Try a W1422:

I x  199 in.4  151 in.4

o.k.

From AISC Manual Table 6-2, the available flexural strength and shear strength for a W1422 is determined as follows. Assume the beam has full lateral support; therefore, Lb = 0. LRFD b M nx  125 kip-ft  91.8 kip-ft o.k. vVn  94.5 kips  12.2 kips o.k.

ASD M nx  82.8 kip-ft  67.5 kip-ft o.k. b

Vn  63.0 kips  9.00 kips o.k. v

Note: The beams and supporting girders in this area should be rechecked when the final weights and locations for the mechanical units have been determined.

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SELECT ROOF BEAMS AT THE END (EAST & WEST) OF THE BUILDING The beams at the ends of the building carry the brick spandrel panel and a small portion of roof load. For these beams, the cladding weight exceeds 25% of the total dead load on the beam. Therefore, per AISC Design Guide 3, limit the vertical deflection due to cladding and initial dead load to L/600 or a in. maximum. In addition, because these beams are supporting brick above and there is continuous glass below, limit the superimposed dead and live load deflection to L/600 or 0.3 in. maximum to accommodate the brick and L/360 or 4 in. maximum to accommodate the glass. Therefore, combining the two limitations, limit the superimposed dead and live load deflection to L/600 or 4 in. The superimposed dead load includes all of the dead load that is applied after the cladding has been installed. In calculating the wall loads, the spandrel panel weight is taken as 55 psf. Beam loading is calculated as follows and shown in Figure III-4. Note, the beams are laterally supported by the deck as shown in Detail 4 on Sheet S4.1.

The dead load from the spandrel is:



wD   7.50 ft  0.055 kip/ft 2



 0.413 kip/ft

The dead load from the roof is equal to:



wD   3.50 ft  0.020 kip/ft 2



 0.070 kip/ft

Use 8 psf for the initial dead load, which includes the deck, beams and joists:



wD (initial )   3.50 ft  0.008 kip/ft 2



 0.028 kip/ft

Use 12 psf for the superimposed dead load:



wD ( super )   3.50 ft  0.012 kip/ft 2



 0.042 kip/ft

The snow load from the roof conservatively uses the maximum snow drift as a uniform load, considering both side and end parapet drift pressures:



wS   3.50 ft  0.025 kip/ft 2  0.0132 kip/ft 2



 0.134 kip/ft

From ASCE/SEI 7, Chapter 2, calculate the required strength of the beams at the east and west ends of the roof. LRFD wu  1.2  0.483 kip/ft   1.6  0.134 kip/ft 

 0.794 kip/ft

ASD wa  0.483 kip/ft  0.134 kip/ft

 0.617 kip/ft

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LRFD 22.5 ft   Ru   0.794 kip/ft     2   8.93 kips

Mu 

ASD 22.5 ft   Ra   0.617 kip/ft     2   6.94 kips

 0.794 kip/ft  22.5 ft 2

Ma 

8

 50.2 kip-ft

 0.617 kip/ft  22.5 ft 2 8

 39.0 kip-ft

Assume the beams are simple spans of 22.5 ft. Calculate the minimum moment of inertia to limit the superimposed dead and live load deflection after cladding is installed to L/600 or ¼ in.

 22.5 ft 12 in./ft  L   4 in. 600 600  0.450 in.  4 in. Therefore, limit deflection to ¼ in. Using the equation for deflection derived previously, the required moment of inertia, Ix req, can be determined as follows:

I x req 

 0.042 kip/ft  0.134 kip/ft  22.5 ft 4 1,290 4 in.

 140 in.4 Calculate minimum moment of inertia to limit the cladding and initial dead load deflection to L/600 or a in.

 22.5 ft 12 in./ft  L   a in. 600 600  0.450 in.  a in. Therefore, limit deflection to a in. Using the equation for deflection derived previously, the required moment of inertia, Ix req, can be determined as follows:

I x req 

 0.413 kip/ft  0.028 kip/ft  22.5 ft 4 1,290  a in.

 234 in.4

controls

Fig. III-4. Beam loading and bracing diagram for roof beams at east and west ends of building.

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From AISC Manual Table 3-3, select a beam size with an adequate moment of inertia. Try a W1626:

I x  301 in.4  234 in.4

o.k.

From AISC Manual Table 6-2, the available flexural strength and shear strength for a W1626 is determined as follows. The beam has full lateral support; therefore, Lb = 0. LRFD b M nx  166 kip-ft  50.2 kip-ft o.k. vVn  106 kips  8.93 kips o.k.

ASD M nx  110 kip-ft  39.0 kip-ft o.k. b

Vn  70.5 kips  6.94 kips o.k. v

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SELECT ROOF BEAMS ALONG THE SIDE (NORTH & SOUTH) OF THE BUILDING

The beams along the side of the building carry the spandrel panel and a substantial roof dead load and live load. For these beams, the cladding weight exceeds 25% of the total dead load on the beam. From AISC Design Guide 3, limit the vertical deflection due to cladding and initial dead load to L/600 or a in. maximum. In addition, because these beams are supporting brick above and there is continuous glass below, limit the superimposed dead and live load deflection to L/600 or 0.3 in. maximum to accommodate the brick and L/360 or 4 in. maximum to accommodate the glass. Therefore, combining the two limitations, limit the superimposed dead and live load deflection to L/600 or 4 in. The superimposed dead load includes all of the dead load that is applied after the cladding has been installed. These beams will be part of the moment frames on the side of the building and therefore will be designed as fixed at both ends. The roof dead load and snow load on this edge beam is equal to the joist end dead load and snow load reaction. Treat this as a uniform load and divide by the joist spacing. (Note: treating this as a uniform load is a convenient and reasonable approximation in this case, resulting in a difference in maximum moment of approximately 4% as compared to the moment calculated using concentrated loading from each of the roof joists acting on the beam). Beam loading is calculated as follows, and shown in Figure III-5. The dead load from the joist end reaction is:

2.76 kips 6.00 ft  0.460 kip/ft

wD 

From previous calculations, the dead load from the spandrel is: wD  0.413 kip/ft

The snow load from the joist end reaction is:

3.73 kips 6.00 ft  0.622 kip/ft

wS 

Use 8 psf for initial dead load and 12 psf for superimposed dead load.









wD (initial )   22.5 ft  0.5 ft  0.008 kip/ft 2  0.184 kip/ft wD ( super )   22.5 ft  0.5 ft  0.012 kip/ft 2  0.276 kip/ft

Fig. III-5. Loading and bracing diagram for roof beams at north and south ends of building.

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From ASCE/SEI 7, Chapter 2, calculate the required strength of the beams at the roof sides. LRFD wu  1.2  0.873 kip/ft   1.6  0.622 kip/ft 

 2.04 kip/ft  30 ft  Ru   2.04 kip/ft     2   30.6 kips

ASD wa  0.873 kip/ft  0.622 kip/ft  1.50 kip/ft  30 ft  Ra  1.50 kip/ft     2   22.5 kips

Using the equation for deflection derived previously, the required moment of inertia, Ix req, is determined as follows. To limit the superimposed dead and live load deflection to 4 in.:

I x req 

 0.622 kip/ft  0.276 kip/ft  30 ft 4 6,440 4 in.

 452 in.4

controls

To limit the cladding and initial dead load deflection to a in.:

I req 

 0.597 kip/ft  30.0 ft 4 6,440  a in.

 200 in.4 From AISC Manual Table 3-3, select a beam size with an adequate moment of inertia. Try a W1835:

I x  510 in.4  452 in.4

o.k.

Calculate Cb for compression in the bottom flange braced at the midpoint and supports using AISC Specification Equation F1-1. Moments along the span are summarized in Figure III-6. LRFD From AISC Manual Table 3-23, Case 15: M u max 

 2.04 kip/ft  30 ft 2

12  153 kip-ft (at supports)

M a max 

1.50 kip/ft  30 ft 2

12  113 kip-ft (at supports)

At midpoint:

At midpoint: Mu 

ASD From AISC Manual Table 3-23, Case 15:

 2.04 kip/ft  30 ft 2

Ma 

1.50 kip/ft  30 ft 2

24  56.3 kip-ft

24  76.5 kip-ft

LRFD

ASD

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At quarter-point of unbraced length:

At quarter-point of unbraced length:

2 2.04 kip/ft  6  30 ft  3.75 ft    30 ft   M uA  12  6  3.75 ft 2     52.6 kip-ft

2 1.50 kip/ft  6  30 ft  3.75 ft    30 ft   M aA  12  6  3.75 ft 2     38.7 kip-ft

At midpoint of unbraced length:

At midpoint of unbraced length:

2 2.04 kip/ft  6  30 ft  7.50 ft    30 ft   12  6  7.50 ft 2     19.1 kip-ft

2 1.50 kip/ft  6  30 ft  7.50 ft    30 ft   12  6  7.50 ft 2     14.1 kip-ft

M uB 

M aB 

At three-quarter point of unbraced length:

At three-quarter point of unbraced length:

2.04 kip/ft  6  30 ft 11.3 ft    30 ft  12  6 11.3 ft 2   62.5 kip-ft

M uC 

Using AISC Specification Equation F1-1:

Cb  

12.5M max 2.5M max  3M A  4M B  3M C 12.5 153 kip-ft 

  

1.50 kip/ft 6  30 ft 11.3 ft    30 ft  12  6 11.3 ft 2   46.0 kip-ft

M aC 

2

  

Using AISC Specification Equation F1-1:

Cb 

 2.5 153 kip-ft   3  52.6 kip-ft       4 19.1 kip-ft   3  62.5 kip-ft  

 2.38

2



12.5M max 2.5M max  3M A  4M B  3M C 12.5 113 kip-ft 

 2.5 113 kip-ft   3  38.7 kip-ft       4 14.1 kip-ft   3  46.0 kip-ft  

 2.38

(a) LRFD

(b) ASD Fig. III-6. Beam moment diagram.

From AISC Manual Table 6-2, with Lb = 6 ft and Cb = 1.0 the available flexural strength is determined as follows:

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LRFD b M n  229 kip-ft  76.5 kip-ft

o.k.

ASD Mn  152 kip-ft  56.3 kip-ft o.k. b

From AISC Manual Table 6-2, with Lb = 15 ft and Cb = 2.38, the available flexural strength is determined as follows:

LRFD

ASD

b M n Cb  b M p

Mp Mn Cb  b b

109 kip-ft  2.38  249 kip-ft

 72.4 kip-ft  2.38  166 kip-ft

259 kip-ft  249 kip-ft

172 kip-ft  166 kip-ft

Therefore:

Therefore:

b M n  249 kip-ft  153 kip-ft

o.k.

Mn  166 kip-ft  113 kip-ft o.k. b

From AISC Manual Table 6-2, the available shear strength is determined as follows:

LRFD vVn  159 kips  30.6 kips o.k.

ASD Vn  106 kips  22.5 kips o.k. v

Therefore, the W1835 is acceptable.

Note: This roof beam may need to be upsized during the lateral load analysis to increase the stiffness and strength of the member and improve lateral frame drift performance.

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SELECT THE ROOF BEAMS ALONG THE INTERIOR LINES OF THE BUILDING

There are three individual beam loadings that occur along grids C and D. The beams from 1 to 2 and 7 to 8 have a uniform snow load except for the snow drift at the end at the parapet. The snow drift from the far ends of the 45-ft joists is negligible. The beams from 2 to 3 and 6 to 7 are the same as the first group, except they have snow drift at the screen wall. The live load deflection is limited to L/240 (or 1.50 in.). Joist reactions are divided by the joist spacing and treated as a uniform load, just as they were for the side beams.  45 ft  30 ft  wD  0.020 kip/ft 2   2    0.750 kip/ft





 45 ft  30 ft  wS  0.025 kip/ft 2   2    0.938 kip/ft





The loading diagrams with moments and end reactions are shown in Figure III-7.

(a) Grids 1 to 2 and 7 to 8

(b) Grids 2 to 3 and 6 to 7 Fig. III-7. Roof beam loading and bracing diagram.

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From ASCE/SEI 7, Chapter 2, the required strength for the beams from grids 1 to 2 and 7 to 8 (opposite hand) is determined as follows: LRFD Ru (left end)  1.2 11.6 kips   1.6 16.0 kips 

ASD Ra (left end)  11.6 kips  16.0 kips

 27.6 kips

 39.5 kips Ru (right end)  1.2 11.2 kips   1.6 14.2 kips   36.2 kips M u  1.2  84.3 kip-ft   1.6 107 kip-ft   272 kip-ft

Ra (right end)  11.2 kips  14.2 kips  25.4 kips M a  84.3 kip-ft  107 kip-ft  191 kip-ft

Using the equation for deflection derived previously, the minimum moment of inertia, Ix req, to limit the live load deflection to 1.50 in., considering a 30-ft simply supported beam and neglecting the modest snow drift is:

 0.938 kip/ft  30 ft 4 I x req  1,290 1.50 in.  393 in.4

From AISC Manual Table 3-3, select a beam size with an adequate moment of inertia. Try a W2144:

I x  843 in.4  393 in.4

o.k.

From AISC Manual Table 6-2, for a W2144 with Lb = 6 ft and Cb = 1.0, the available flexural strength and shear strength is determined as follows:

LRFD b M n  332 kip-ft  272 kip-ft

ASD Mn  221 kip-ft  191 kip-ft o.k. b

o.k.

Vn  145 kips  27.6 kips o.k. v

vVn  217 kips  39.5 kips o.k.

From ASCE/SEI 7, Chapter 2, the required strength for the beams from grids 2 to 3 and 6 to 7 (opposite hand) is determined as follows: LRFD Ru (left end)  1.2 11.3 kips   1.6 14.4 kips 

 25.7 kips

 36.6 kips Ru (right end)  1.2 11.3 kips   1.6 17.9 kips 

Ra (right end)  11.3 kips  17.9 kips  29.2 kips

 42.2 kips M u  1.2  84.4 kip-ft   1.6 111 kip-ft   279 kip-ft

ASD Ra (left end)  11.3 kips  14.4 kips

M a  84.4 kip-ft  111 kip-ft  195 kip-ft

From AISC Manual Table 6-2, for a W2144 with Lb = 6 ft and Cb = 1.0, the available flexural strength and shear strength is determined as follows:

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LRFD b M n  332 kip-ft  279 kip-ft

ASD Mn  221 kip-ft  195 kip-ft o.k. b

o.k.

Vn  145 kips  29.2 kips o.k. v

vVn  217 kips  42.2 kips o.k.

The third individual beam loading occurs at the beams from 3 to 4, 4 to 5, and 5 to 6. For these beams there is a uniform snow load outside the screen walled area, except for the snow drift at the parapet ends and the screen wall ends of the 45-ft-long joists. Inside the screen walled area the beams support the mechanical equipment. The loading diagram is shown in Figure III-8.

 2.70 kips  2  15 ft  wD      0.360 kip/ft   6 ft   6 ft   1.35 kip/ft





 4.02 kips  2  15 ft  wS      0.240 kip/ft   6 ft   6 ft   1.27 kip/ft





From ASCE/SEI 7, Chapter 2, the required strength for the beams from grids 3 to 4, 4 to 5, and 5 to 6 is determined as follows: LRFD wu  1.2 1.35 kip/ft   1.6 1.27 kip/ft 

 2.62 kip/ft

 3.65 kip/ft

Mu 

ASD wa  1.35 kip/ft  1.27 kip/ft

 3.65 kip/ft  30 ft 2

Ma 

 2.62 kip/ft  30 ft 2

8  295 kip-ft

8

 411 kip-ft

Fig. III-8. Loading and bracing diagram for roof beams from grid 3 to 4, 4 to 5, and 5 to 6.

LRFD

ASD

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 30 ft  Ru   3.65 kip/ft     2   54.8 kips

 30 ft  Ra   2.62 kip/ft     2   39.3 kips

Using the equation for deflection derived previously, the minimum moment of inertia, Ix req, to limit the live load deflection to 1.50 in. is:

1.27 kip/ft  30 ft 4 I x req  1,290 1.50 in.  532 in.4

From AISC Manual Table 3-3, select a beam size with an adequate moment of inertia. Try a W2155:

I x  1,140 in.4  532 in.4

o.k.

From AISC Manual Table 6-2, for a W2155 with Lb = 6 ft and Cb = 1.0, the available flexural strength and shear strength is determined as follows:

LRFD b M n  473 kip-ft  411 kip-ft

o.k.

vVn  234 kips  54.8 kips o.k.

ASD Mn  314 kip-ft  295 kip-ft o.k. b

Vn  156 kips  39.3 kips o.k. v

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FLOOR MEMBER DESIGN AND SELECTION

Calculate dead load and live load. Dead load: Slab and deck Beams (est.) Misc. (ceiling, mechanical, etc.) Total

= 57 psf = 8 psf = 10 psf = 75 psf

Note: The weight of the floor slab and deck was obtained from the manufacturer’s literature. Live load: Total (can be reduced for area per ASCE/SEI 7) = 80 psf The floor and deck will be 3 in. of normal weight concrete, f c = 4 ksi, on 3-in., 20 gage, galvanized, composite deck, laid in a pattern of three or more continuous spans. The total depth of the slab is 6 in. From the Steel Deck Institute Floor Deck Design Manual (SDI, 2014), the maximum unshored span for construction with this deck and a three-span condition is 10 ft 6 in. The general layout for the floor beams is 10 ft on center; therefore, the deck does not need to be shored during construction. At 10 ft on center, this deck has an allowable superimposed live load capacity of 143 psf. In addition, it can be shown that this deck can carry a 2,000 pound load over an area of 2.5 ft by 2.5 ft as required by ASCE/SEI 7, Section 4.4. The floor diaphragm and the floor loads extend 6 in. past the centerline of grid as shown on Sheet S4.1.

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SELECT FLOOR BEAMS (COMPOSITE AND NONCOMPOSITE)

Note: There are two early and important checks in the design of composite beams. First, select a beam that either does not require camber, or establish a target camber and moment of inertia at the start of the design process. A reasonable approximation of the camber is between L/300 minimum and L/180 maximum (or a maximum of 12 to 2 in.). Second, check that the beam is strong enough to safely carry the wet concrete and a 20 psf construction live load [per Design Loads on Structures During Construction, ASCE 37-14 (ASCE, 2014)] when designed by the ASCE/SEI 7 load combinations and the provisions of AISC Specification Chapter F. SELECT TYPICAL 45-FT-LONG INTERIOR COMPOSITE BEAM (10 FT ON CENTER)

Find a target moment of inertia for an unshored beam.



wD  10 ft  0.057 kip/ft 2  0.008 kip/ft 2



 0.650 kip/ft Hold deflection to 2 in. maximum to facilitate concrete placement. Using the equation for deflection derived previously, the required moment of inertia is determined as follows:

 0.650 kip/ft  45 ft 4 I req  1,290  2 in.  1, 030 in.4

The construction live load is determined as follows:



wL  10 ft  0.020 kip/ft 2



 0.200 kip/ft

From ASCE/SEI 7, the required flexural strength due to wet concrete only is determined as follows: wu  1.4  0.650 kip/ft 

LRFD

ASD wa  0.650 kip/ft

 0.910 kip/ft

Mu 

 0.910 kip/ft  45 ft 2

Ma 

8

 230 kip-ft

 0.650 kip/ft  45 ft 2 8

 165 kip-ft

From ASCE/SEI 7, the required flexural strength due to wet concrete and construction live load is determined as follows: LRFD wu  1.2  0.650 kip/ft   1.6  0.200 kip/ft   1.10 kip/ft

ASD wa  0.650 kip/ft  0.200 kip/ft

 0.850 kip/ft

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LRFD Mu 

1.10 kip/ft  45 ft 

ASD

2

Ma 

8  278 kip-ft controls

 0.850 kip/ft  45 ft 2 8

 215 kip-ft

controls

Use AISC Manual Table 3-2 to select a beam with Ix  1,030 in.4 Select W2150, with Ix = 984 in.4, close to the target value. From AISC Manual Table 6-2, the available flexural strength for a fully braced, Lb = 0 ft, W2150 is determined as follows: LRFD b M n  413 kip-ft  278 kip-ft

ASD

Mn  274 kip-ft  215 kip-ft o.k. b

o.k.

Check for possible live load reduction due to area in accordance with ASCE/SEI 7, Section 4.7.2. From ASCE/SEI 7, Table 4.7-1, for interior beams: K LL  2

The beams are at 10 ft on center, therefore the tributary area is: AT   45 ft 10 ft   450 ft 2



K LL AT  2 450 ft 2  900 ft



2

Because KLLAT  400 ft2, a reduced live load can be used.

From ASCE/SEI 7, Equation 4.7-1:  15 L  Lo  0.25  K LL AT 

   0.50Lo 

 15   80 psf   0.25   900 ft 2   60.0 psf  40.0 psf

   0.50  80 psf   

Therefore, use L = 60.0 psf. The beams are at 10 ft on center, therefore the loading is as shown in Figure III-9. Note, the beam is continuously braced by the deck.

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Back III-24

Fig. III-9. Loading and bracing diagram for typical interior composite floor beams. From ASCE/SEI 7, Chapter 2, the required strengths are determined as follows: LRFD wu  1.2  0.750 kip/ft   1.6  0.600 kip/ft   1.86 kip/ft

 45 ft  Ra  1.35 kip/ft     2   30.4 kips

 45 ft  Ru  1.86 kip/ft     2   41.9 kips

Mu 

ASD wa  0.750 kip/ft  0.600 kip/ft  1.35 kip/ft

1.86 kip/ft  45 ft 2

Ma 

1.35 kip/ft  45 ft 2

8  342 kip-ft

8

 471 kip-ft

The available flexural strength for the composite beam is determined using AISC Manual Part 3. Assume initially a = 1 in. Y 2  Ycon 

a 2

 6.00 in. 

(Manual Eq. 3-6) 1 in. 2

 5.50 in.

Enter AISC Manual Table 3-19 for a W2150 with Y2 = 5.50 in. Selecting PNA location 7, with Qn = 184 kips, the available flexural strength is:

LRFD b M n  598 kip-ft  471 kip-ft o.k.

ASD Mn  398 kip-ft  342 kip-ft o.k. b

Determine effective width, b The effective width of the concrete slab is the sum of the effective widths for each side of the beam centerline as determined by the minimum value of the three widths set forth in AISC Specification Section I3.1a:

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Back III-25

1. one-eighth of the span of the beam, center-to-center of the supports  45 ft     2 sides   11.3 ft  8 

2. one-half the distance to the centerline of the adjacent beam  10 ft     2 sides   10.0 ft  2 

controls

3. distance to the edge of the slab The latter is not applicable for an interior member. Determine the height of the compression block, a.

a 

 Qn 0.85 f c b

(Manual Eq. 3-7)

184 kips 0.85  4 ksi 10 ft 12 in./ft 

 0.451 in.  1.00 in. o.k. From AISC Manual Table 6-2, the available shear strength of the W2150 bare steel beam is determined as follows: LRFD vVn  237 kips  41.9 kips

o.k.

ASD Vn  158 kips  30.4 kips o.k. v

Check live load deflection

 45 ft 12 in./ft  L  360 360  1.50 in. Entering AISC Manual Table 3-20 for a W2150, with PNA location 7 and Y2 = 5.50 in., provides a lower bound moment of inertia of ILB = 1,730 in.4 From the equation previously derived, the live load deflection is determined as follows:  LL  

wL L4 1, 290 I LB

 0.600 kip/ft  45 ft 4



1, 290 1, 730 in.4



 1.10 in.  1.50 in. o.k.

From AISC Design Guide 3 limit the live load deflection, using 50% of the (unreduced) design live load, to L/360 with a maximum absolute value of 1 in. across the bay. From the equation previously derived, the deflection is determined as follows:

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 LL 

0.5  0.800 kip/ft  45 ft 



1, 290 1, 730 in.4

4



 1 in.

 0.735 in.  1 in.  0.735 in. 1 in.  0.735 in.  0.265 in.

Note: Limit the supporting girders to 0.265 in. deflection under the same load case at the connection point of the beam. Determine the required number of shear stud connectors From AISC Manual Table 3-21, using perpendicular deck with one w-in.-diameter anchor per rib in normal weight concrete with fc = 4 ksi in the weak position: Qn  17.2 kips/anchor

Qn Qn 184 kips  17.2 kips/anchor  10.7 anchors (on each side of maximum moment)

n

Therefore, 22 studs are required to satisfy strength requirements. However, per AISC Specification Commentary Section I3.2d.1, 44 studs are specified to provide sufficient deformation capacity by ensuring a degree of composite action of at least 50%. From AISC Design Guide 3, limit the wet concrete deflection in a bay to L/360, not to exceed 1 in. From the equation previously derived, the wet concrete deflection is determined as follows:  DL ( wet conc ) 

 0.650 kip/ft  45 ft 4



1, 290 984 in.4



 2.10 in.

Camber the beam for 80% of the calculated wet deflection.

Camber  0.80  2.10 in.

 1.68 in. Round the calculated value down to the nearest 4 in.; therefore, specify 12 in. of camber. 2.10 in.  12 in.  0.600 in. 1 in.  0.600 in.  0.400 in.

Note: Limit the supporting girders to 0.400 in. deflection under the same load combination at the connection point of the beam.

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SELECT TYPICAL 30-FT INTERIOR COMPOSITE (OR NONCOMPOSITE) BEAM (10 FT ON CENTER) Find a target moment of inertia for an unshored beam. Determine the required strength to carry wet concrete and construction live load. The dead load from the slab and deck is:



wD  10 ft  0.057 kip/ft 2  0.008 kip/ft 2



 0.650 kip/ft

Hold deflection to 12 in. maximum to facilitate concrete placement. Using the equation for deflection derived previously, the required moment of inertia is determined as follows: I req 

 0.650 kip/ft  30 ft 4 1,290 12 in.

 272 in.4

The construction live load is:



wL  10 ft  0.020 kip/ft 2



 0.200 kip/ft

From ASCE/SEI 7, Chapter 2, determine the required flexural strength due to wet concrete only. wu  1.4  0.650 kip/ft 

LRFD

ASD wa  0.650 kip/ft

 0.910 kip/ft

Mu 

 0.910 kip/ft  30 ft 2

Ma 

8

 102 kip-ft

 0.650 kip/ft  30 ft 2 8

 73.1 kip-ft

From ASCE/SEI 7, Chapter 2, determine the required flexural strength due to wet concrete and construction live load.

LRFD wu  1.2  0.650 kip/ft   1.6  0.200 kip/ft   1.10 kip/ft

Mu 

1.10 kip/ft  30 ft 2

 124 kip-ft

8 controls

ASD wa  0.650 kip/ft  0.200 kip/ft  0.850 kip/ft Ma 

 0.850 kip/ft  30 ft 2

8  95.6 kip-ft controls

Use AISC Manual Table 3-2 to find a beam with an Ix  272 in.4 Select W1626, with Ix = 301 in.4, which exceeds the target value.

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From AISC Manual Table 6-2, the available flexural strength for a fully braced, Lb = 0 ft, W1626 is determined as follows: LRFD b M n  166 kip-ft  124 kip-ft

ASD Mn  110 kip-ft  95.6 kip-ft o.k. b

o.k.

Check for possible live load reduction due to area in accordance with ASCE/SEI 7, Section 4.7.2. From ASCE/SEI 7, Table 4.7-1, for interior beams: K LL  2

The beams are at 10 ft on center, therefore the tributary area is: AT   30 ft 10 ft   300 ft 2



K LL AT  2 300 ft 2



 600 ft 2

Because KLLAT  400 ft2, a reduced live load can be used.

From ASCE/SEI 7, Equation 4.7-1:  15 L  Lo  0.25  K LL AT 

   0.50Lo 

 15   80 psf   0.25   600 ft 2   69.0 psf  40.0 psf

   0.50  80 psf   

Therefore, use L = 69.0 psf. The beams are at 10 ft on center, therefore the loading is as shown in Figure III-10.

Fig. III-10. Loading and bracing diagram for typical 30-ft interior floor beams.

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From ASCE/SEI 7, Chapter 2, calculate the required strength. LRFD wu  1.2  0.750 kip/ft   1.6  0.690 kip/ft   2.00 kip/ft

 30 ft  Ra  1.44 kip/ft     2   21.6 kips

 30 ft  Ru   2.00 kip/ft     2   30.0 kips

Mu 

ASD wa  0.750 kip/ft  0.690 kip/ft  1.44 kip/ft

 2.00 kip/ft  30 ft 2

Ma 

8  225 kip-ft

1.44 kip/ft  30 ft 2 8

 162 kip-ft

The available flexural strength for the composite beam is determined from AISC Manual Part 3 as follows. Assume initially that a = 1 in. Y 2  Ycon 

a 2

 6.00 in. 

(Manual Eq. 3-6) 1 in. 2

 5.50 in.

Enter AISC Manual Table 3-19 for a W1626 with Y2 = 5.50 in. Selecting PNA location 7, with Qn = 96.0 kips, the available flexural strength is:

LRFD

ASD

Mn  165 kip-ft  162 kip-ft o.k. b

b M n  248 kip-ft  225 kip-ft o.k.

Determine effective width, b The effective width of the concrete slab is the sum of the effective widths for each side of the beam centerline as determined by the minimum value of the three widths set forth in AISC Specification Section I3.1a: 1. one-eighth of the span of the beam, center-to-center of the supports  30 ft     2 sides   7.50 ft  8 

controls

2. one-half the distance to the centerline of the adjacent beam  10 ft     2 sides   10.0 ft  2 

3. distance to the edge of the slab The latter is not applicable for an interior member. Determine the height of the compression block, a.

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Back III-30

a 

 Qn 0.85 fc b

(Manual Eq. 3-7)

96.0 kips 0.85  4 ksi  7.50 ft 12 in./ft 

 0.314 in.  1.00 in. o.k. From AISC Manual Table 6-2, the available shear strength of the W1626 bare steel beam is determined as follows: LRFD vVn  106 kips  30.0 kips

ASD

Vn  70.5 kips  21.6 kips o.k. v

o.k.

Check live load deflection

 30 ft 12 in./ft  L  360 360  1.00 in. Entering AISC Manual Table 3-20 for a W1626, with PNA location 7 and Y2 = 5.50 in., provides a lower bound moment of inertia of ILB = 575 in.4 From the equation previously derived, the live load deflection is determined as follows:  LL  

wL L4 1, 290 I LB

 0.690 kip/ft  30 ft 4



1, 290 575 in.4



 0.753 in.  1.00 in.

o.k.

From AISC Design Guide 3, limit the live load deflection, using 50% of the (unreduced) design live load, to L/360 with a maximum absolute value of 1 in. across the bay. From the equation previously derived, the deflection is determined as follows:

 LL 

0.5  0.800 kip/ft  30 ft 



1, 290 575 in.4

4



 0.437 in.  1 in. o.k. 1 in.  0.437 in.  0.563 in.

Note: Limit the supporting girders to 0.563 in. deflection under the same load combination at the connection point of the beam. Determine the required number of shear stud connectors From AISC Manual Table 3-21, using perpendicular deck with one w-in.-diameter anchor per rib in normal weight concrete with fc = 4 ksi in the weak position: Qn  17.2 kips/anchor

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Qn Qn 96.0 kips  17.2 kips/anchor  5.58 anchors (on each side of maximum moment)

n

Note: Per AISC Specification Section I8.2d, there is a maximum spacing limit of 8(6 in.) = 48 in. (not to exceed 36 in.) between anchors.

Therefore use 12 anchors, uniformly spaced at no more than 36 in. on center. Per AISC Specification Commentary Section I3.2d.1, beams with spans not exceeding 30 ft are not susceptible to connector failure due to insufficient connector capacity. Note: Although the studs may be placed up to 36 in. on center, the steel deck must still be anchored to the supporting member at a spacing not to exceed 18 in. per AISC Specification Section I3.2c. From AISC Design Guide 3, limit the wet concrete deflection in a bay to L/360, not to exceed 1 in. From the equation previously derived, the wet concrete deflection is determined as follows:  DL ( wet conc ) 

 0.650 kip/ft  30 ft 4



1, 290 301 in.4



 1.36 in.

Camber the beam for 80% of the calculated wet concrete dead load deflection.

Camber  0.80 1.36 in.

 1.09 in. Round the calculated value down to the nearest 4 in. Therefore, specify 1 in. of camber. 1.36 in.  1 in.  0.360 in.

1.00 in.  0.360 in.  0.640 in.

Note: Limit the supporting girders to 0.640 in. deflection under the same load combination at the connection point of the beam. This beam could also be designed as a noncomposite beam. Try a W1835. From AISC Manual Table 6-2 the available flexural strength for a fully braced beam, Lb = 0 ft, and shear strength are determined as follows. LRFD b M n  249 kip-in.  225 kip-ft o.k.

vVn  159 kips  30.0 kips o.k.

ASD Mn  166 kip-in.  162 kip-ft o.k. b

Vn  106 kips  21.6 kips o.k. v

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Check beam deflections Check live load deflection. From AISC Manual Table 3-2 for a W1835: Ix = 510 in.4  LL 

 0.690 kip/ft  30 ft 4



1, 290 510 in.4

 0.850 in. < 1 in.



o.k.

Based on AISC Design Guide 3, limit the live load deflection, using 50% of the (unreduced) design live load, to L/360 with a maximum absolute value of 1 in. across the bay. From the equation previously derived, the deflection is determined as follows:

 LL 

0.5  0.800 kip/ft  30 ft 

4

1, 290  510 in.4 

 0.492 in.  1 in. o.k. 1 in.  0.492 in.  0.508 in.

Note: Limit the supporting girders to 0.508 in. deflection under the same load combination at the connection point of the beam. Note: Because this beam is stronger than the W1626 composite beam, no wet concrete strength checks are required in this example. From AISC Design Guide 3, limit the wet concrete deflection in a bay to L/360, not to exceed 1 in. From the equation previously derived, the wet concrete deflection is determined as follows:  DL ( wet conc ) 

 0.650 kip/ft  30 ft 

4

1, 290  510 in.4 

 0.800 in.  1 in. o.k.

Camber the beam for 80% of the calculated wet concrete deflection.

Camber  0.80  0.800 in.

 0.640 in. A good break point to eliminate camber is w in.; therefore, do not specify a camber for this beam. 1 in.  0.800 in.  0.200 in.

Note: Limit the supporting girders to 0.200 in. deflection under the same load case at the connection point of the beam. Therefore, selecting a W1835 will eliminate both shear studs and cambering. The cost of the extra steel weight may be offset by the elimination of studs and cambering. Local labor and material costs should be checked to make this determination.

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SELECT TYPICAL NORTH-SOUTH EDGE BEAM The influence area, KLLAT, for these beams is less than 400 ft2; therefore, no live load reduction can be taken per ASCE/SEI 7, Section 4.7.2. These beams carry 5.5 ft of dead load and live load as well as a wall load. The floor dead load is:



w   5.5 ft  0.075 kip/ft 2



 0.413 kip/ft

Use 65 psf for the initial dead load due to the wet concrete:



wD (initial )   5.5 ft  0.065 kip/ft 2



 0.358 kip/ft

Use 10 psf for the superimposed dead load:



wD ( super )   5.5 ft  0.010 kip/ft 2



 0.055 kip/ft

The dead load of the wall system at the floor is:







w   7.50 ft  0.055 kip/ft 2   6.00 ft  0.015 kip/ft 2



 0.413 kip/ft  0.090 kip/ft  0.503 kip/ft

The total dead load is:

wD  0.413 kip/ft  0.503 kip/ft

 0.916 kip/ft The live load is:



wL   5.5 ft  0.080 kip/ft 2



 0.440 kip/ft

Beam loading is shown in Figure III-11.

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Fig. III-11. Loading and bracing diagram for typical north-south floor beams. Calculate the required strengths from ASCE/SEI 7, Chapter 2: LRFD wu  1.2  0.916 kip/ft   1.6  0.440 kip/ft 

 1.80 kip/ft

 22.5 ft  Ra  1.36 kip/ft     2   15.3 kips

 22.5 ft  Ru  1.80 kip/ft     2   20.3 kips

Mu 

ASD wa  0.916 kip/ft  0.440 kip/ft  1.36 kip/ft

1.80 kip/ft  22.5 ft 2

Ma 

8

1.36 kip/ft  22.5 ft 2 8

 86.1 kip-ft

 114 kip-ft

Because these beams are less than 25 ft long, they will be most efficient as noncomposite beams. The beams at the edges of the building carry a brick spandrel panel. For these beams, the cladding weight exceeds 25% of the total dead load on the beam. From AISC Design Guide 3, limit the vertical deflection due to cladding and initial dead load to L/600 or a in. maximum. In addition, because these beams are supporting brick above and there is continuous glass below, limit the superimposed dead and live load deflection to L/600 or 0.3 in. maximum to accommodate the brick and L/360 or 4 in. maximum to accommodate the glass. Therefore, combining the two limitations, limit the superimposed dead and live load deflection to L/600 or 4 in. The superimposed dead load includes all of the dead load that is applied after the cladding has been installed. Note that it is typically not recommended to camber beams supporting spandrel panels. Using the equation for deflection derived previously, the minimum moment of inertia, Ix superimposed dead and live load deflection to 4 in. I x req 

req,

to limit the

 0.055 kip/ft  0.440 kip/ft  22.5 ft 4 1, 290 4 in.

 393 in.4

Using the equation for deflection derived previously, the minimum moment of inertia, Ix req, to limit the cladding and initial dead load deflection to a in. I x req 

 0.358 kip/ft  0.503 kip/ft  22.5 ft 4 1, 290  a in.

 456 in.4

controls

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From AISC Manual Table 3-2, find a beam with Ix  456 in.4 Select a W1835 with Ix = 510 in.4

From AISC Manual Table 6-2, the available flexural strength for a fully braced beam, Lb = 0 ft, and shear strength are determined as follows: LRFD

ASD

b M n  249 kip-in.  114 kip-ft o.k.

vVn  159 kips  20.3 kips o.k.

Mn  166 kip-in.  86.1 kip-ft o.k. b

Vn  106 kips  15.3 kips o.k. v

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SELECT TYPICAL EAST-WEST EDGE GIRDER The beams along the sides of the building carry the spandrel panel and glass, and dead load and live load from the intermediate floor beams. For these beams, the cladding weight exceeds 25% of the total dead load on the beam. Therefore, per AISC Design Guide 3, limit the vertical deflection due to cladding and initial dead load to L/600 or a in. maximum. In addition, because these beams are supporting brick above and there is continuous glass below, limit the superimposed dead and live load deflection to L/600 or 0.3 in. maximum to accommodate the brick and L/360 or 4 in. maximum to accommodate the glass. Therefore, combining the two limitations, limit the superimposed dead and live load deflection to L/600 or 4 in. The superimposed dead load includes all of the dead load that is applied after the cladding has been installed. These beams will be part of the moment frames on the north and south sides of the building and therefore will be designed as fixed at both ends.

Establish the loading. The dead load reaction from the floor beams is:  45 ft  PD   0.750 kip/ft     2   16.9 kips  45 ft  PD (initial )   0.650 kip/ft     2   14.6 kips  45 ft  PD ( super )   0.100 kip/ft     2   2.25 kips The uniform dead load along the beam is:





wD   0.5 ft  0.075 kip/ft 2  0.503 kip/ft  0.541 kip/ft



wD (initial )   0.5 ft  0.065 kip/ft 2



 0.033 kip/ft



wD ( super )   0.5 ft  0.010 kip/ft 2



 0.005 kip/ft

Select typical 30-ft composite (or noncomposite) girders. Check for possible live load reduction due to area in accordance with ASCE/SEI 7, Section 4.7.2. From ASCE/SEI 7, Table 4.7-1, for edge beams with cantilevered slabs: K LL  1

However, it is also permissible to calculate the value of KLL based upon influence area. Because the cantilever dimension is small, KLL will be closer to 2 than 1. The calculated value of KLL based upon the influence area is:

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K LL 

 45.5 ft  30 ft 

 45 ft   0.5 ft   30 ft    2   1.98

AT   30 ft  22.5 ft  0.5 ft   690 ft 2

From ASCE/SEI 7, Equation 4.7-1:  L  Lo  0.25  

   0.50Lo K LL AT  15

 15    80 psf   0.25  1.98 690 ft 2    52.5 psf  40.0 psf





    0.50  80 psf   

Therefore, use L = 52.5 psf. The live load from the floor beams is:

 45 ft  PL   0.525 kip/ft     2   11.8 kips The uniform live load along the beam is:



wL   0.5 ft  0.0526 kip/ft 2



 0.0263 kip/ft

The loading diagram is shown in Figure III-12.

Fig. III-12. Loading and bracing diagram for typical east-west edge girders.

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The required moment and end reactions at the floor side beams are determined from a structural analysis of a fixedend beam and summarized as follows: LRFD

ASD

Typical side beam:

Typical side beam:

Ru  49.5 kips

Ra  37.2 kips

M u (at ends)  313 kip-ft

M a (at ends)  234 kip-ft

M u (at center)  156 kip-ft

M a (at center)  117 kip-ft

The maximum moment occurs at the support with compression in the bottom flange. The bottom flange is laterally braced at 10 ft on center by the intermediate beams. Note: During concrete placement, because the deck is parallel to the beam, the beam will not have continuous lateral support. It will be braced at 10 ft on center by the intermediate beams. By inspection, this condition will not control because the maximum moment under full loading causes compression in the bottom flange, which is braced at 10 ft on center. ASD

LRFD Calculate Cb for compression in the bottom flange braced at 10 ft on center.

Calculate Cb for compression in the bottom flange braced at 10 ft on center.

Cb = 2.21 (from computer output)

Cb = 2.22 (from computer output)

Select a W2144.

Select a W2144.

With continuous bracing, Lb = 0 ft, from AISC Manual Table 6-2:

With continuous bracing, Lb = 0 ft, from AISC Manual Table 6-2:

b M n  358 kip-ft  156 kip-ft

Mn  238 kip-ft  117 kip-ft b

o.k.

o.k.

From AISC Manual Table 6-2 with Lb = 10 ft and Cb = 2.21:

From AISC Manual Table 6-2 with Lb = 10 ft and Cb = 2.22:

b M n Cb   264 kip-ft  2.21

Mn Cb  176 kip-ft  2.22  b  391 kip-ft

 583 kip-ft

From AISC Specification Section F2.2, the nominal flexural strength is limited to Mp.

From AISC Specification Section F2.2, the nominal flexural strength is limited to Mp.

b M n  b M p

Mn M p  b b 391 kip-ft  238 kip-ft

583 kip-ft  358 kip-ft

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ASD

LRFD Therefore:

Therefore: b M n  358 kip-ft  313 kip-ft

Mn  238 kip-ft  234 kip-ft o.k. b

o.k.

From AISC Manual Table 6-2, the available shear strength is determined as follows: LRFD vVn  217 kips  49.5 kips

ASD Vn  145 kips  37.2 kips o.k. v

o.k.

Deflections are determined from a structural analysis of a fixed-end beam. For deflection due to cladding and initial dead load:   0.295 in.  a in.

o.k.

For deflection due to superimposed dead and live loads:   0.212 in.  4 in.

o.k.

Note that both of the deflection criteria stated previously for the girder and for the locations on the girder where the floor beams are supported have also been met. Also, as noted previously, it is not typically recommended to camber beams supporting spandrel panels. The W2144 is adequate for strength and deflection, but may be increased in size to help with moment frame strength or drift control.

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SELECT TYPICAL EAST-WEST INTERIOR GIRDER Establish loads The dead load reaction from the floor beams is:  45 ft  30 ft  PD   0.750 kip/ft    2    28.1 kips Check for live load reduction due to area in accordance with ASCE/SEI 7, Section 4.7.2. From ASCE/SEI 7, Table 4.7-1, for interior beams:

KLL = 2  45 ft  30 ft  AT   30 ft    2    1,130 ft 2 Using ASCE/SEI 7, Equation 4.7-1:  L  Lo  0.25  

   0.50 Lo K LL AT  15

 15    80 psf   0.25    2  1,130 ft 2   45.2 psf  40.0 psf





    0.50  80 psf   

Therefore, use L = 45.2 psf. The live load from the floor beams is:  45 ft  30 ft  PL  0.0452 kip/ft 2   10 ft  2    17.0 kips





The loading is shown in Figure III-13.

Fig. III-13. Loading and bracing diagram for typical interior girder.

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Note: The dead load for this beam is included in the assumed overall dead load. From ASCE/SEI 7, Chapter 2, the required strengths are determined as follows: LRFD

Ru  1.2  28.1 kips   1.6 17.0 kips   60.9 kips

ASD

Ra  28.1 kips  17.0 kips  45.1 kips M a   45.1 kips 10 ft 

M u   60.9 kips 10 ft 

 451 kip-ft

 609 kip-ft

Check for beam requirements when carrying wet concrete. Limit wet concrete deflection to 12 in.

 45 ft  30 ft  PD   0.650 kip/ft    2    24.4 kips  45 ft  30 ft  PL   0.200 kip/ft    2    7.50 kips Note: During concrete placement, because the deck is parallel to the beam, the beam will not have continuous lateral support. It will be braced at 10 ft on center by the intermediate beams. Also, during concrete placement, a construction live load of 20 psf will be present. The loading is shown in Figure III-14. From ASCE/SEI 7, Chapter 2, the required strengths for the typical interior beams with wet concrete only is determined as follows: Ru  1.4  24.4 kips   34.2 kips M u   34.2 kips 10 ft 

LRFD

ASD

Ra  24.4 kips M a   24.4 kips 10 ft   244 kip-ft

 342 kip-ft

Fig. III-14. Loading and bracing diagram for typical interior girder with wet concrete and construction loads.

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From ASCE/SEI 7, Chapter 2, the required strengths for the typical interior beams with wet concrete and construction live load is determined as follows: LRFD

ASD

Ru  1.2  24.4 kips   1.6  7.50 kips 

Ra  24.4 kips  7.50 kips  31.9 kips

 41.3 kips

M a (midspan)   31.9 kips 10 ft 

M u (midspan)   41.3 kips 10 ft 

 319 kip-ft

 413 kip-ft

Assume Ix  935 in.4, which is determined based on a wet concrete deflection of 12 in. From AISC Manual Table 3-2, select a W2168 with Ix = 1,480 in.4. From AISC Manual Table 6-2, verify the available flexural strength and shear strength using Lb = 10 ft, and Cb = 1.0. LRFD

ASD

b M n  532 kip-ft  413 kip-ft o.k.

Mn  354 kip-ft  319 kip-ft o.k. b

vVn  272 kips  41.3 kips o.k.

Vn  181 kips  31.9 kips o.k. v

Check W2168 as a composite beam.

From previous calculations: LRFD

ASD

Ru  60.9 kips

Ra  45.1 kips

M u (midspan)  609 kip-ft

M a (midspan)  451 kip-ft

From previous calculations, assuming a = 1 in.: Y 2  5.50 in.

Enter AISC Manual Table 3-19 for a W2168 with Y2 = 5.50 in. Selecting PNA location 7 with Qn = 250 kips provides an available flexural strength of:

LRFD b M n  844 kip-ft  609 kip-ft o.k.

ASD

Mn  561 kip-ft  451 kip-ft o.k. b

From AISC Design Guide 3, limit the wet concrete deflection in a bay to L/360, not to exceed 1 in. From AISC Manual Table 3-23, Case 9:

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 DL ( wet conc ) 

23PL L3 648 EI 23  24.4 kips  30 ft  12 in./ft  3





648  29, 000 ksi  1, 480 in.4

3



 0.941 in. Camber the beam for 80% of the calculated wet concrete deflection.

Camber  0.80  0.941 in.

 0.753 in. Round the calculated value down to the nearest 4 in,: therefore, specify w-in. of camber. 0.941 in.  w in.  0.191 in.  0.400 in.

Therefore, the total deflection limit of 1 in. for the bay has been met. Determine the effective width, b From AISC Specification Section I3.1a, the effective width of the concrete slab is the sum of the effective widths for each side of the beam centerline, which shall not exceed: 1. one-eighth of the span of the beam, center-to-center of supports

 30 ft     2 sides   7.50 ft controls  8  2. one-half the distance to the centerline of the adjacent beam

 45 ft 30 ft      37.5 ft 2   2 3. the distance to the edge of the slab The latter is not applicable for an interior member. Determine the height of the compression block, a a 

 Qn 0.85 f c b

(Manual Eq. 3-7)

250 kips 0.85  4 ksi  7.50 ft 12 in./ft 

 0.817 in.  1 in. o.k.

From AISC Manual Table 6-2, the available shear strength of the W2168 is determined as follows. LRFD

vVn  272 kips  60.9 kips o.k.

ASD

Vn  181 kips  45.1 kips o.k. v

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Check live load deflection.

 LL  

L 360  30 ft 12 in./ft 

360  1.00 in. Entering AISC Manual Table 3-20 for a W2168, with PNA location 7 and Y2 = 5.50 in., provides a lower bound moment of inertia of ILB = 2,510 in.4  LL 

23PL L3 648 EI LB 23 17.0 kips  30 ft  12 in./ft  3



3



648  29, 000 ksi  2,510 in.4

 0.387 in.  1.00 in.



o.k.

From AISC Design Guide 3, limit the live load deflection, using 50% of the (unreduced) design live load, to L/360 with a maximum absolute value of 1 in. across the bay. The maximum deflection is: 23  0.5  30.0 kips  30 ft  12 in./ft  3

 LL 



648  29, 000 ksi  2,510 in.4

3



 0.341 in.  1.00 in. o.k.

Check the deflection at the location where the floor beams are supported.  LL 

0.5  30.0 kips 120 in.  2 3  360 in.120 in.  4 120 in.  4   6  29, 000 ksi  2,510 in.



 0.297 in.  0.265 in.



o.k.

Therefore, the total deflection in the bay is 0.297 in. + 0.735 in. = 1.03 in., which is acceptably close to the limit of 1 in, where LL = 0.735 in. is from the 45 ft interior composite beam running north-south.

Determine the required shear stud connectors Using Manual Table 3-21, for parallel deck with, wr/hr  1.5, one w-in.-diameter stud in normal weight, 4-ksi concrete: Qn = 21.5 kips/anchor  Qn 250 kips  Qn 21.5 kips/anchor  11.6 anchors (on each side of maximum moment)

Therefore, use a minimum of 24 studs for horizontal shear. Per AISC Specification Section I8.2d, the maximum stud spacing is 36 in.

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Since the load is concentrated at 3 points, the studs are to be arranged as follows: Use 12 studs between supports and supported beams at third points. Between supported beams (middle third of span), use 4 studs to satisfy minimum spacing requirements. Therefore, 28 studs are required in a 12:4:12 arrangement. Notes: Although the studs may be placed up to 36 in. on center, the steel deck must still be anchored to be the supporting member at a spacing not to exceed 18 in. in accordance with AISC Specification Section I3.2c. This W2168 beam, with full lateral support, is very close to having sufficient available strength to support the imposed loads without composite action. A larger noncomposite beam might be a better solution.

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COLUMN DESIGN AND SELECTION FOR GRAVITY LOADS Estimate column loads

Roof loads (from previous calculations): Dead load Snow load Total

= 20 psf = 25 psf = 45 psf

The snow drift loads at the perimeter of the roof and at the mechanical screen wall are developed from previous calculations. Reaction to column (side parapet):



 3.73 kips  2 w   0.025 kip/ft  6.00 ft   0.0467 kip/ft

  23.0 ft 

where 3.73 kips is the snow load reaction, including drift, from the 24KCS4 roof joist at the side parapet. Reaction to column (end parapet):





 16.0 kips  2 w   0.025 kip/ft 15.5 ft   37.5 ft   0.0392 kip/ft where 16.0 kips is the snow load reaction, including drift, from the W2144 roof beam along the interior lines of the building. Reaction to column (screen wall along lines C & D):



 4.02 kips  2 w   0.025 kip/ft 6 ft    0.108 kip/ft

  22.5 ft 

where 4.02 kips is the snow load reaction, including drift, from the 24KCS4 joist at the screen wall. Mechanical equipment and screen wall (average): w = 40 psf The spandrel panel weight was calculated as 0.413 kip/ft as part of the selection process for the W1626 roof beams at the east and west ends of the building. The mechanical room dead load of 0.060 kip/ft2 and snow load of 0.040 kip/ft2 was determined as part of the selection process for the W1422 roof beams at the mechanical area.

A summary of the column loads at the roof is given in Table III-2.

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Column 2A, 2F, 3A, 3F, 4A, 4F, 5A, 5F, 6A ,6F, 7A, 7F Snow drifting at side Exterior wall

Table III-2 Summary of Column Loads at the Roof Loading Area, DL, PD, Width, Length, ft ft ft2 kip/ft2 kips 23.0 30.0 690 0.020 13.8 30.0 30.0

1B, 1E, 8B, 8E Snow drifting at side Exterior wall

3.50

1A, 1F, 8A, 8F

23.0

0.413 kip/ft

22.5 22.5 22.5

78.8

15.5

357

0.020 0.413 kip/ft 0.020

12.4 26.2 1.58 9.29 10.9 6.36

SL,

PS,

kip/ft2 0.025

kips 17.3

0.0467 kip/ft

1.40

0.025 0.0392 kip/ft

18.7 1.97 0.882

0.025

2.85 7.95

0.0392 kip/ft 0.0467 kip/ft

0.463 0.724

0.025

9.14 13.6

0.0392 kip/ft

1.03

0.025 0.025

14.6 28.1 16.9

78.8 ft 2 2 = 318 

Snow drifting at end Snow drifting at side Exterior wall 1C, 1D, 8C, 8D

11.8 15.5 27.3 37.5

15.5

0.413 kip/ft 581

0.020

11.3 17.7 10.8

78.8 ft 2 2 = 542 

Snow drifting at end Exterior wall 2C, 2D, 7C, 7D 3C, 3D, 4C, 4D, 5C, 5D, 6C, 6D Snow drifting Mechanical area

26.3 26.3

0.413 kip/ft

37.5 22.5

30.0 30.0

1,125 675

0.020 0.020

15.0

30.0 30.0

450

0.060

10.9 21.7 22.5 13.5 27.0 40.5

0.108 kip/ft 0.040

3.24 18.0 38.1

Floor loads (from previous calculations): Dead load Snow load Total

= 75 psf = 80 psf = 155 psf

Calculate reduction in live loads, analyzed at the base of three floors (n = 3) using ASCE/SEI 7, Section 4.7.2. Note that the 6-in. cantilever of the floor slab has been ignored for the calculation of KLL for columns in this building because it has a negligible effect. Columns:

2A, 2F, 3A, 3F, 4A, 4F, 5A, 5F, 6A, 6F, 7A, 7F Exterior column without cantilever slabs KLL = 4 (ASCE/SEI 7, Table 4.7-1) Lo = 80 psf n = 3 (three floors supported)

AT   22.5 ft  0.5 ft  30 ft   690 ft 2

Using ASCE/SEI 7, Equation 4.7-1:

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 15 L  Lo  0.25  K LL nAT 

   0.4 Lo 

 15    80 psf   0.25+  4  3 690 ft 2   33.2 psf  32.0 psf





    0.4  80 psf  

Therefore, use L = 33.2 psf. Columns:

1B, 1E, 8B, 8E Exterior column without cantilever slabs KLL = 4 (ASCE/SEI 7, Table 4.7-1) Lo = 80 psf n=3

AT   5.00 ft  0.5 ft  22.5 ft   124 ft 2

 15 L  Lo  0.25  K LL nAT      80 psf   0.25+ 

   0.4 Lo  15

 4  3 124 ft 2 

    0.4  80 psf  

 51.1 psf  32.0 psf

Use L = 51.1 psf. Columns: 1A, 1F, 8A, 8F Corner column without cantilever slabs KLL = 4 (ASCE/SEI 7, Table 4.7-1) Lo = 80 psf n=3  124 ft 2 AT  15.0 ft  0.5 ft  22.5 ft  0.5 ft     2

  

 295 ft 2  15 L  Lo  0.25  K LL nAT 

   0.4 Lo 

 15    80 psf  0.25+  4  3 295 ft 2   40.2 psf  32.0 psf





    0.4  80 psf  

Therefore, use L = 40.2 psf.

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Columns: 1C, 1D, 8C, 8D Exterior column without cantilever slabs KLL = 4 (ASCE/SEI 7, Table 4.7-1) Lo = 80 psf n=3 2  45 ft  30 ft   124 ft AT  15.0 ft  0.5 ft    –  2    2

  

 519 ft 2  15 L  Lo  0.25  K LL nAT 

   0.4 Lo 

 15    80 psf   0.25   4  3 519 ft 2   35.2 psf  32.0 psf





    0.4  80 psf  

Therefore, use L = 35.2 psf. Columns: 2C, 2D, 3C, 3D, 4C, 4D, 5C, 5D, 6C, 6D, 7C, 7D Interior column KLL = 4 (ASCE/SEI 7, Table 4.7-1) Lo = 80 psf n=3  45 ft  30 ft  AT     30 ft  2    1,125 ft 2

 15 L  Lo  0.25  K LL nAT     80 psf  0.25+  

   0.4 Lo 

   0.4  80 psf  2   4  3 1,125 ft   15

 30.3 psf  32.0 psf Therefore, use L = 32.0 psf. A summary of the column loads at the floors is given in Table III-3.

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Column

2A, 2F, 3A, 3F, 4A, 4F, 5A, 5F, 6A ,6F, 7A, 7F Exterior wall

Table III-3 Summary of Column Loads at the Floors Loading Area, DL, PD, Width, Length, ft ft ft2 kip/ft2 kips 23.0

30.0

690

30.0

0.075

51.8

0.503 kip/ft

15.1 66.9 9.30 11.3 20.6 22.1

1B, 1E, 8B, 8E Exterior wall

5.50

22.5 22.5

124

0.075 0.503 kip/ft

1A, 1F, 8A, 8F

23.0

15.5

357

0.075

27.3

124 ft 2 2  295 0.503 kip/ft

15.5

581

LL,

PL ,

kip/ft2

kips

0.0332

22.9

0.0511

22.9 6.34

0.0402

6.34 11.9

0.0352

11.9 18.3

 Exterior wall 1C, 1D, 8C, 8D

37.5

0.075

124 ft 2 2  519 0.503 kip/ft

13.7 35.8 38.9

 Exterior wall 2C, 2D, 3C, 3D, 4C, 4D, 5C, 5D, 6C, 6D, 7C, 7D

26.3 37.5

30.0

1,125

0.075

13.2 52.1 84.4

18.3 0.0320

36.0

The spandrel panel weight was calculated as 0.503 kip/ft as part of the selection process for the W1835 edge beams at the north and south ends of the building. The column loads are summarized in Table III-4.

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Column 2A, 2F, 3A, 3F, 4A, 4F, 5A, 5F, 6A, 6F, 7A, 7F

1B, 1E, 8B, 8E

1A, 1F, 8A, 8F

1C, 1D, 8C, 8D

2C, 2D, 7C, 7D

3C, 3D, 4C, 4D, 5C, 5D, 6C, 6D

Table III-4 Summary of Column Loads Floor PD, kips Roof 26.2 4th 66.9 3rd 66.9 2nd 66.9 Total 227 Roof 10.9 4th 20.6 3rd 20.6 2nd 20.6 Total 72.7 Roof 17.7 4th 35.8 3rd 35.8 2nd 35.8 Total 125 Roof 21.7 4th 52.1 3rd 52.1 2nd 52.1 Total 178 Roof 22.5 4th 84.4 3rd 84.4 2nd 84.4 Total 276 Roof 40.5 4th 84.4 3rd 84.4 2nd 84.4 Total 294

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PL , kips 22.9 22.9 22.9 68.7 6.34 6.34 6.34 19.0 11.9 11.9 11.9 35.7 18.3 18.3 18.3 54.9 36.0 36.0 36.0 108 36.0 36.0 36.0 108

PS, kips 18.7

18.7 2.85

2.85 9.14

9.14 14.6

14.6 28.1

28.1 38.1

38.1

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SELECT TYPICAL INTERIOR LEANING COLUMNS Columns 3C, 3D, 4C, 4D, 5C, 5D, 6C, 6D

Elevation of second floor slab: Elevation of first floor slab: Column unbraced length:

113.5 ft 100 ft Lx = Ly = 13.5 ft

Note: Kx = Ky = 1.0 for a leaning column when using the effective length method. Lcx  K x Lx  1.0 13.5 ft   13.5 ft Lcy  K y Ly  1.0 13.5 ft   13.5 ft From ASCE/SEI 7, Chapter 2, the required axial strength is determined using the following controlling load combinations (including the 0.5 live load reduction permitted for LRFD): LRFD

Pu  1.2  294 kips   1.6 108 kips   0.5  38.1 kips 

ASD

Pa  294 kips  0.75 108 kips   0.75  38.1 kips   404 kips

 545 kips

Using AISC Manual Table 4-1a, enter with Lc = 14.0 ft (conservative) and proceed across the table until reaching the lightest size that has sufficient available strength at the required unbraced length. Select a W1265. The available strength in axial compression is: LRFD

c Pn  685 kips  545 kips

ASD

Pn  456 kips  404 kips o.k. c

o.k.

Note: A W1468 would also be an acceptable selection. Columns 2C, 2D, 7C, 7D

Elevation of second floor slab: 113.5 ft Elevation of first floor slab: 100 ft Column unbraced length: Lx = Ly = 13.5 ft Note: Kx = Ky = 1.0 for a leaning column when using the effective length method. Lcx  K x Lx  1.0 13.5 ft   13.5 ft Lcy  K y Ly  1.0 13.5 ft   13.5 ft

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From ASCE/SEI 7, Chapter 2, the required axial strength is determined using the following controlling load combinations (including the 0.5 live load reduction permitted for LRFD): LRFD

Pu  1.2  276 kips   1.6 108 kips   0.5  28.1 kips 

ASD

Pa  276 kips  0.75 108 kips   0.75  28.1 kips   378 kips

 518 kips

Using AISC Manual Table 4-1a, enter with Lc = 14.0 ft (conservative) and proceed across the table until reaching the lightest size that has sufficient available strength at the required unbraced length. Select a W1265. The available strength in axial compression is: LRFD

c Pn  685 kips  518 kips

ASD

Pn  456 kips  378 kips o.k. c

o.k.

Note: A W1461 would also be an acceptable selection. However, W1265 columns were selected to keep sizes consistent for all interior columns. SELECT TYPICAL EXTERIOR LEANING COLUMNS Columns 1B, 1E, 8B, 8E

Elevation of second floor slab: 113.5 ft Elevation of first floor slab: 100 ft Column unbraced length: Lx = Ly = 13.5 ft Note: Kx = Ky = 1.0 for a leaning column when using the effective length method. Lcx  K x Lx  1.0 13.5 ft   13.5 ft Lcy  K y Ly  1.0 13.5 ft   13.5 ft From ASCE/SEI 7, Chapter 2, the required axial strength is determined using the following controlling load combinations (including the 0.5 live load reduction permitted for LRFD): LRFD

Pu  1.2  72.7 kips   1.6 19.0 kips   0.5  2.85 kips 

ASD

Pa  72.7 kips  0.75 19.0 kips   0.75  2.85 kips   89.1 kips

 119 kips

Using AISC Manual Table 4-1a, enter with Lc = 14.0 ft (conservative) and proceed across the table until reaching the lightest size that has sufficient available strength at the required unbraced length. Select a W1240. The available strength in axial compression is: LRFD

c Pn  304 kips  119 kips

o.k.

ASD

Pn  202 kips  89.1 kips o.k. c

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Note, A W12 column was selected above for ease of erection of framing beams (bolted double-angle connections can be used without bolt staggering). Final column selections at the moment and braced frames are illustrated later in this example.

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WIND LOAD DETERMINATION Use the Envelope Procedure for simple diaphragm buildings from ASCE/SEI 7, Chapter 28, Part 2. To qualify for the simplified wind load method for low-rise buildings, per ASCE/SEI 7, Section 28.5.2, the following conditions must be met: 1.

Simple diaphragm building o.k.

2.

Low-rise building ≤ 60 ft o.k.

3.

Enclosed, and conforms to wind borne debris provisions o.k.

4.

Regular-shaped o.k.

5.

Not a flexible building o.k.

6.

Does not have response characteristics requiring special considerations o.k.

7.

Symmetrical shape with flat or gable roof with  ≤ 45º o.k.

8.

Torsional load cases from ASCE/SEI 7, Figure 28.3-1 do not control design of MWFRS o.k.

Define input parameters 1.

Risk category:

II from ASCE/SEI 7, Table 1.5-1

2.

Basic wind speed:

V = 107 mph (3-s) from ASCE/SEI 7, Figure 26.5-1B

3.

Exposure category:

C from ASCE/SEI 7, Section 26.7.3

4.

Topographic factor:

Kzt = 1.0 from ASCE/SEI 7, Section 26.8.2

5.

Mean roof height:

55.0 ft

6.

Height and exposure adjustment:

1.59 from ASCE/SEI 7, Figure 28.5-1

7.

Roof angle:

 = 0

ps  K zt ps 30

(ASCE/SEI 7, Eq. 28.5-1)

 1.59 1.0 18.2 psf   28.9 psf (Horizontal pressure zone A)  1.59 1.0 12.0 psf   19.1 psf (Horizontal pressure zone C)  1.59 1.0  21.9 psf   34.8 psf (Vertical pressure zone E)  1.59 1.0  12.4 psf   19.7 psf (Vertical pressure zone F)  1.59 1.0  15.2 psf   24.2 psf (Vertical pressure zone G)  1.59 1.0  9.59 psf   15.2 psf (Vertical pressure zone H) a = 10% of least horizontal dimension or 0.4h, whichever is smaller, but not less than either 4% of least horizontal dimension or 3 ft

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a = the lesser of: 10% of the least horizontal dimension = 12.3 ft 40% of the eave height = 22.0 ft but not less than 4% of the least horizontal dimension or 3 ft = 4.92 ft Thus, a = 12.3 ft and 2a = 24.6 ft. Zone A: End zone of wall (width = 2a) Zone C: Interior zone of wall Zone E: End zone of windward roof (width = 2a) Zone F: End zone of leeward roof (width = 2a) Zone G: Interior zone of windward roof Zone H: Interior zone of leeward roof Calculate load on roof diaphragm Mechanical screen wall height: Wall height: Parapet wall height: Total wall height at roof at screen wall: Total wall height at roof at parapet:

6 ft 0.5[55 ft – 3(13.5 ft)] = 7.25 ft 2 ft 6 ft  7.25 ft  13.3 ft 2 ft  7.25 ft  9.25 ft

ws ( A)   28.9 psf  9.25 ft   267 plf ws (C )  19.1 psf  9.25 ft   177 plf (at parapet) ws (C )  19.1 psf 13.3 ft   254 plf (at screen wall)

Calculate load on fourth floor diaphragm

0.5  55.0 ft  40.5 ft   7.25 ft

Wall height:

0.5  40.5 ft  27.0 ft   6.75 ft 6.75 ft  7.25 ft  14.0 ft

Total wall height at floor: ws ( A)   28.9 psf 14.0 ft   405 plf ws (C )  19.1 psf 14.0 ft   267 plf

Calculate load on third floor diaphragm Wall height:

0.5  40.5 ft  27.0 ft   6.75 ft 0.5  27.0 ft  13.5 ft   6.75 ft

Total wall height at floor:

6.75 ft  6.75 ft  13.5 ft

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ws ( A)   28.9 psf 13.5 ft   390 plf ws (C )  19.1 psf 13.5 ft   258 plf

Calculate load on second floor diaphragm

0.5  27.0 ft  13.5 ft   6.75 ft

Wall height:

0.5 13.5 ft  0 ft   6.75 ft 6.75 ft  6.75 ft  13.5 ft

Total wall height at floor: ws ( A)   28.9 psf 13.5 ft   390 plf ws (C )  19.1 psf 13.5 ft   258 plf

Determine the wind load on each frame at each level. Conservatively apply the end zone pressures on both ends of the building simultaneously, where l = length of structure, ft b = width of structure, ft h = height of wall at building element, ft For wind from a north or south direction: Total load to each frame:

l  PW  N -S   ws A  2a   ws C    2a  2   Shear in diaphragm: v N -S  

v N -S  

PW  N -S  120 ft

PW  N -S  90 ft

, for roof

, for floors (deduction for stair openings)

For wind from an east or west direction: Total load to each frame:

b  PW  E -W   ws A  2a   ws C    2a  2   Shear in diaphragm:

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v E -W  

PW  E -W  210 ft

, for roof and floors

Table III-5 summarizes the total wind load in each direction acting on a steel frame at each level. The wind load at the ground level has not been included in the chart because it does not affect the steel frame. The roof level dimensions exclude the screen wall area. The floor level dimensions correspond to the outside dimensions of the cladding.

l, Screen Roof 4th 3rd 2nd Base

ft 93.0 120 213 213 213

b, ft 33.0 90.0 123 123 123

2a, ft 0 24.6 24.6 24.6 24.6

Table III-5 Summary of Wind Loads at Each Level ps(C), ws(A), ws(C), PW(N-S), h, ps(A), ft psf psf plf plf kips 13.3 0 19.1 0 254 11.8 9.25 28.9 19.1 267 177 12.8 14.0 28.9 19.1 405 267 31.8 13.5 28.9 19.1 390 258 30.7 13.5 28.9 19.1 390 258 30.7 118

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PW(E-W), kips 4.19 10.2 19.8 19.1 19.1 72.4

v(N-S), plf  205 353 341 341

v(E-W), plf  68.5 94.3 91.0 91.0

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SEISMIC LOAD DETERMINATION

The floor plan area: 120 ft, column center line to column center line, by 210 ft, column centerline to column center line, with the edge of floor slab or roof deck 6 in. beyond the column center line.

Area  121 ft  211 ft   25,500 ft 2 The perimeter cladding system length: Length   2 123 ft    2  213 ft   672 ft

The perimeter cladding weight at floors:

 7.50 ft   0.055 kip/ft 2 

Brick spandrel panel with metal stud backup:

 6.00 ft   0.015 kip/ft 2 

Window wall system: Total:

= 0.413 kip/ft = 0.090 kip/ft 0.503 kip/ft

Typical roof dead load (from previous calculations): Although 40 psf was used to account for the mechanical units and screen wall for the beam and column design, the entire mechanical area will not be uniformly loaded. Use 30% of the uniform 40 psf mechanical area load to determine the total weight of all of the mechanical equipment and screen wall for the seismic load determination. Roof area:

 25, 500 ft  0.020 kip/ft 

Wall perimeter:

 672 ft  0.413 kip/ft 

2

2

= 278 kips

 2, 700 ft   0.3  0.040 kip/ft  2

Mechanical area:

= 510 kips 2

= 32.4 kips 820 kips

Total:

Typical third and fourth floor dead load: Note: An additional 10 psf has been added to the floor dead load to account for partitions per ASCE/SEI 7, Section 12.7.2. Floor area:

 25, 500 ft  0.085 kip/ft 

= 2,170 kips

Wall perimeter:

 672 ft  0.503 kip/ft 

= 338 kips

2

2

Total:

2,510 kips

Second floor dead load (the floor area is reduced because of the open atrium): Floor area:

 24, 700 ft  0.085 kip/ft 

= 2,100 kips

Wall perimeter:

 672 ft  0.503 kip/ft 

= 338 kips

Total:

2

2

2,440 kips

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Total dead load of the building: Roof Fourth floor Third floor Second floor Total

820 kips 2,510 kips 2,510 kips 2,440 kips 8,280 kips

Calculate the seismic forces. Determine the seismic risk category and importance factors. Office Building: Risk Category II, from ASCE/SEI 7, Table 1.5-1 Seismic Importance Factor: Ie = 1.00, from ASCE/SEI 7, Table 1.5-2 The site coefficients are given in this example. SS and S1 can also be determined from ASCE/SEI 7, Figures 22-1 and 22-2, respectively.

SS = 0.121g S1 = 0.060g

Soil, Site Class D (given) Fa @ SS M 0.25 = 1.6 from ASCE/SEI 7, Table 11.4-1 Fv @ S1 M 0.1 = 2.4 from ASCE/SEI 7, Table 11.4-2

Determine the maximum considered earthquake accelerations. From ASCE/SEI 7, Equation 11.4-1: S MS  Fa S S  1.6  0.121g   0.194 g From ASCE/SEI 7, Equation 11.4-2: S M 1  Fv S1  2.4  0.060 g   0.144 g Determine the design earthquake accelerations. From ASCE/SEI 7, Equation 11.4-3: S DS  qS MS  q  0.194 g   0.129 g From ASCE/SEI 7, Equation 11.4-4:

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S D1  qS M 1  q  0.144 g   0.096 g Determine the seismic design category from ASCE/SEI 7, Table 11.6-1. With SDS < 0.167g and Risk Category II, Seismic Design Category A applies. With 0.067g M SD1 < 0.133g and Risk Category II, Seismic Design Category B applies.

Select the seismic force-resisting system from ASCE/SEI 7, Table 12.2-1. For Seismic Design Category B it is permissible to select a structural steel system not specifically detailed for seismic resistance (Item H). The response modification coefficient, R, is 3. Determine the approximate fundamental period, Ta. Building height, hn = 55.0 ft

Ct = 0.02 and x = 0.75 from ASCE/SEI 7, Table 12.8-2 (“All other structural systems”) From ASCE/SEI 7, Equation 12.8-7: Ta  Ct hnx   0.02  55.0 ft 

(ASCE/SEI 7, Eq. 12.8-7) 0.75

 0.404 s

Determine the redundancy factor from ASCE/SEI 7, Section 12.3.4.1.  = 1.0, for Seismic Design Category B From ASCE/SEI 7, Equation 12.4-4a, determine the vertical seismic effect term:

Ev  0.2 S DS D

(ASCE/SEI 7, Eq. 12.4-4a)

 0.2  0.129 g  D  0.0258 D From ASCE/SEI 7, Equation 12.4-3, determine the horizontal seismic effect term:

Eh  QE

(ASCE/SEI 7, Eq. 12.4-3)

 1.0  QE  The following seismic load combinations are as specified in ASCE/SEI 7, Sections 2.3.6 and 2.4.5 as directed by Section 12.4.2. Where the prescribed seismic load effect, E = f(Ev, Eh), is combined with the effects of other loads, the following load combinations apply. Note that L = 0.5L for LRFD per ASCE/SEI 7, Section 2.3.6 Exception 1. LRFD 1.2 D  Ev  Eh  L  0.2S  1.2 D  0.2S DS D  QE  0.5 L  0.2S  1.2  0.0258  D  1.0QE  0.5 L  0.2 S  1.23D  1.0QE  0.5 L  0.2S

ASD

1.0 D  0.7 Ev  0.7 Eh  1.0 D  0.7  0.2S DS D   0.7QE  1.0  0.7  0.0258   D  0.7 1.0  QE  1.02 D  0.7QE

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LRFD 0.9 D  Ev  Eh  0.9 D  0.2S DS D  QE

ASD 1.0 D  0.525Ev  0.525Eh  0.75L  0.75S  1.0 D  0.525  0.2S DS D   0.525QE  0.75L  0.75S

  0.9  0.0258  D  1.0QE

 1.0  0.525  0.0258   D  0.525 1.0  QE  0.75L

 0.874 D  1.0QE

 0.75S  1.01D  0.525QE  0.75L  0.75S

0.6 D  0.7 Ev  0.7 Eh  0.6 D  0.7  0.2S DS D   0.7QE  0.6  0.7  0.0258   D  0.7 1.0  QE  0.582 D  0.7QE Where the prescribed seismic load effect with overstrength, E = f(Ev, Emh), is combined with the effects of other loads, the following load combinations apply. The overstrength factor, o, is determined from ASCE/SEI 7, Table 12.2-1. o = 3 for steel systems not specifically detailed for seismic resistance, excluding cantilever column systems. Determine the horizontal seismic effect term including overstrength.

Emh  o QE  Ecl

(from ASCE/SEI 7, Eq. 12.4-7)

 3  QE  where QE is the effect from seismic forces from seismic base shear, V, as calculated per ASCE/SEI 7, Section 12.8.1; diaphragm design forces, Fpx, as calculated per ASCE/SEI 7, Section 12.10; or seismic design force, Fp, as calculated per Section 13.3.1. The capacity-limited horizontal seismic load effect, Ecl, is defined in ASCE/SEI 7, Section 11.3. LRFD 1.2 D  Ev  Emh  L  0.2S  1.2 D  0.2 S DS D  o QE  0.5 L  0.2 S  1.2  0.0258  D  3QE  0.5L  0.2S  1.23D  3.0QE  0.5 L  0.2 S 0.9 D  Ev  Emh  0.9 D  0.2 S DS D  o QE   0.9  0.0258  D  3QE  0.874 D  3.0QE

ASD

1.0 D  0.7 Ev  0.7 Emh  1.0 D  0.7  0.2S DS D   0.7o QE  1.0  0.7  0.0258   D  0.7  3 QE  1.02 D  2.1QE 1.0 D  0.525 Ev  0.525Emh  0.75L  0.75S  1.0 D  0.525  0.2 S DS D   0.525o QE  0.75 L  0.75S  1.0  0.525  0.0258   D  0.525  3 QE  0.75 L  0.75S  1.01D  1.58QE  0.75L  0.75S

0.6 D  0.7 Ev  0.7 Emh  0.6 D  0.7  0.2S DS D   0.7o QE  0.6  0.7  0.0258   D  0.7  3 QE  0.582 D  2.1QE Calculate the seismic base shear using ASCE/SEI 7, Section 12.8.1.

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Determine the seismic response coefficient, Cs, from ASCE/SEI 7, Equation 12.8-2: Cs 

S DS R I   e

0.129  3     1.00   0.0430 

Let Ta = T, as is permitted in Section 12.8.2. From ASCE/SEI 7, Figure 22-14, TL = 12 > T (midwestern city); therefore, use ASCE/SEI 7, Section 12.8.1.1, to determine the upper limit of Cs.

Cs 

S D1 R T   Ie 

(ASCE/SEI 7, Eq. 12.8-3)

0.096  3  0.404    1.00   0.0792 

Cs shall not be taken less than: Cs  0.044S DS I e  0.01

(ASCE/SEI 7, Eq. 12.8-5)

 0.044  0.129 1.00   0.01  0.00568  0.01 Therefore, Cs = 0.0430. Calculate the seismic base shear from ASCE/SEI 7, Section 12.8.1: V  CsW

(ASCE/SEI 7, Eq. 12.8-1)

 0.0430  8, 280 kips   356 kips Determine vertical distribution of seismic forces from ASCE/SEI 7, Section 12.8.3.

Fx  CvxV

(ASCE/SEI 7, Eq. 12.8-11)

 Cvx  356 kips  Cvx 

wx hx k n

 wi hi

(ASCE/SEI 7, Eq. 12.8-12) k

i 1

for structures having a period of 0.5 s or less, k = 1.

Determine horizontal shear distribution at each level per ASCE/SEI 7, Section 12.8.4.

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n

Vx   Fi

(ASCE/SEI, Eq. 12.8-13)

i x

Determine the overturning moment at each level per ASCE/SEI 7, Section 12.8.5. n

M x   Fi ( hi  hx ) i x

The seismic forces at each level are summarized in Table III-6.

wx , kips 820 2,510 2,510 2,440 8,280

Roof 4th 3rd 2nd Base

Table III-6 Summary of Seismic Forces at Each Level Fx, hxk, wxhxk, Cvx ft kip-ft kips 55.0 45,100 0.182 64.8 40.5 102,000 0.411 146 27.0 67,800 0.273 97.2 13.5 32,900 0.133 47.3 248,000 355

Vx, kips 64.8 211 308 355

Mx , kip-ft 940 3,790 7,940 12,700

Calculate strength and determine rigidity of diaphragms. Determine the diaphragm design forces from ASCE/SEI 7, Section 12.10.1.1. Fpx is the largest of: 1. The force Fx at each level determined by the vertical distribution above n

F

i

2. Fpx 

ix n

 wi ix

w px  0.4 S DS I e w px , from ASCE/SEI 7, Equations 12.10-1 and 12.10-3  0.4  0.129 1.00  wpx  0.0516 wpx

3. Fpx  0.2 S DS I e wpx , from ASCE/SEI 7, Equation 12.10-2  0.2  0.129 1.00  wpx  0.0258wpx

The diaphragm shear forces include the effects of openings in the diaphragm (such as stair shafts) and an accidental torsion calculated using an eccentricity of 5% of the building dimension per ASCE/SEI 7, Section 12.8.4. The accidental torsion resulted in a 10% increase in the shear force. A summary of the diaphragm forces is given in Table III-7,

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where Fpx = max(A, B, C) A = force at a level based on the vertical distribution of seismic forces n

= Fpx 

B

 Fi i x n

 wi

w px  0.4 S DS I e w px

i x

C L V

= 0.2 S DS I e w px = the length of the frame connected to the diaphragm (in the N-S or E-W direction) = shear force in the diaphragm

Roof 4th 3rd 2nd

wpx, kips 820 2,510 2,510 2,440

A, kips 64.8 146 97.2 47.3

Table III-7 Summary of Diaphragm Forces B, C, Fpx, L(N-S), kips kips kips ft 42.3 21.2 64.8 240 130 64.8 146 180 130 64.8 130 180 105 63.0 105 180

L(E-W), ft 420 420 420 420

v(N-S), plf 297 892 794 642

v(E-W), plf 170 382 340 275

Roof Roof deck: Support fasteners: Sidelap fasteners: Joist spacing: Diaphragm length: Diaphragm width:

12-in.-deep, 22 gage, wide rib s-in. puddle welds in 36/5 pattern (3) #10 TEK screws s = 6.00 ft 210 ft lv =120 ft

By inspection, the critical condition for the diaphragm is loading from the north or south directions. LRFD From the ASCE/SEI 7 load combinations for strength design, the earthquake load is:

ASD From the ASCE/SEI 7 load combinations for allowable stress design, the earthquake load is:

vr  Eh  QE

vr  0.7 Eh  0.7QE

 1.0  0.297 klf 

 0.7 1.0  0.297 klf 

 0.297 klf

 0.208 klf

The wind load is:

The wind load is:

vr  1.0W

vr  0.6W

 1.0  0.205 klf 

 0.6  0.205 klf 

 0.205 klf

 0.123 klf

From the SDI Diaphragm Design Manual (SDI, 2015), the available shear strengths are determined as follows:

For panel buckling strength: vn = 3.88 klf For connection strength: vn = 0.815 klf

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LRFD

ASD

Panel buckling strength:

Panel buckling strength:

vn  0.80  3.88 klf 

vn 3.88 klf   2.00  1.94 klf  0.208 klf o.k.

 3.10 klf  0.297 ksf o.k.

Connection strength:

Connection strength:

Earthquake

Earthquake

vn  0.55  0.815 klf 

vn 0.815 klf   3.00  0.272 klf  0.208 ksf o.k.

 0.448 klf  0.297 ksf o.k.

Wind

Wind

vn  0.70  0.815 klf 

vn 0.815 klf  2.35   0.347 klf  0.123 ksf o.k.

 0.571 klf  0.205 ksf o.k.

Check diaphragm flexibility. From the SDI Diaphragm Design Manual (SDI, 2015): Dxx  607 ft K1  0.286 ft 1 K 2  870 kip/in. K 4  3.55

From SDI Diaphragm Design Manual, Section 9: K2 0.3Dxx K4   3K1 s s 870 kip/in.  0.3  607 ft   0.286   3 3.55    6.00 ft  6.00 ft  ft   22.3 kip/in.

G 

Seismic loading on diaphragm.

64.8 kips 210 ft  0.309 klf

w

Calculate the maximum diaphragm deflection.

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

wL2 8lv G 

 0.309 klf  210 ft 2  8 120 ft  22.3 kip/in.  0.637 in. Story drift = 0.154 in. (from computer output) The diaphragm deflection exceeds two times the story drift; therefore, the diaphragm may be considered to be flexible in accordance with ASCE/SEI 7, Section 12.3.1.3. The roof diaphragm is flexible in the N-S direction, but using a rigid diaphragm distribution is more conservative for the analysis of this building. By similar reasoning, the roof diaphragm will also be treated as a rigid diaphragm in the E-W direction. Third and fourth floors Floor deck: 3-in.-deep, 22 gage, composite deck with normal weight concrete Support fasteners: s-in. puddle welds in a 36/4 pattern Sidelap fasteners: (3) #10 TEK screws Beam spacing: s = 10 ft Diaphragm length: 210 ft Diaphragm width: 120 ft lv = 120 ft  30 ft = 90 ft, to account for the stairwell By inspection, the critical condition for the diaphragm is loading from the north or south directions. LRFD From the ASCE/SEI 7 load combinations for strength design, the earthquake load for the fourth floor is:

ASD From the ASCE/SEI 7 load combinations for strength design, the earthquake load for the fourth floor is:

vr  Eh  QE

vr  Eh  0.7QE

 1.0  0.892 klf 

 0.7 1.0  0.892 klf 

 0.892 klf

 0.624 klf

For the fourth floor, the wind load is:

For the fourth floor, the wind load is:

vr  1.0W

vr  0.6W

 1.0  0.353 klf 

 0.6  0.353 klf 

 0.353 klf

 0.212 klf

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LRFD From the ASCE/SEI 7 load combinations for strength design, the earthquake load for the third floor is:

ASD From the ASCE/SEI 7 load combinations for strength design, the earthquake load for the third floor is:

vr  Eh

vr  Eh  0.7QE

 QE  1.0  0.794 klf 

 0.7 1.0  0.794 klf 

 0.794 klf

 0.556 klf

For the third floor, the wind load is:

For the third floor, the wind load is:

vr  1.0W

vr  0.6W

 1.0  0.341 klf 

 0.6  0.341 klf 

 0.341 klf

 0.205 klf

From the SDI Diaphragm Design Manual (SDI, 2015), the nominal connection shear strength is vn = 5.38 klf.

Calculate the available strengths. LRFD Connection Strength (same for earthquake or wind) (SDI, 2015)

ASD Connection Strength (same for earthquake or wind) (SDI, 2015)

vn  0.5  5.38 klf 

vn 5.38 klf  3.25   1.66 klf  0.624 klf o.k.

 2.69 klf  0.892 klf o.k.

Check diaphragm flexibility. From the SDI Diaphragm Design Manual (SDI, 2015): K1  0.318 ft 1 K 2  870 kip/in. K 3  2,380 kip/in. K 4  3.54

K2   G     K3 3 K  K s 1   4 870 kip/in.      2,380 kip/in.  0.318   3.54  3   10 ft     ft     2, 450 kip/in.

Fourth floor Calculate seismic loading on the diaphragm based on the fourth floor seismic load.

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146 kips 210 ft  0.695 klf

w

Calculate the maximum diaphragm deflection on the fourth floor. 

wL2 8lv G 

 0.695 klf  210 ft  8  90 ft  2, 450 kip/in. 2



 0.0174 in.

Third floor Calculate seismic loading on the diaphragm based on the third floor seismic load. 130 kips 210 ft  0.619 klf

w

Calculate the maximum diaphragm deflection on the third floor. 

wL2 8lv G 

 0.619 klf  210 ft  8  90 ft  2, 450 kip/in. 2



 0.0155 in.

The diaphragm deflection at the third and fourth floors is less than two times the story drift (story drift = 0.268 in. from computer output); therefore, the diaphragm is considered rigid in accordance with ASCE/SEI 7, Section 12.3.1.3. By inspection, the floor diaphragm will also be rigid in the E-W direction. Second floor Floor deck: Support fasteners: Sidelap fasteners: Beam spacing: Diaphragm length: Diaphragm width:

3-in.-deep, 22 gage, composite deck with normal weight concrete s-in. puddle welds in a 36/4 pattern (3) #10 TEK screws s = 10 ft 210 ft 120 ft

Because of the atrium opening in the floor diaphragm, an effective diaphragm depth of 75 ft will be used for the deflection calculations. By inspection, the critical condition for the diaphragm is loading from the north or south directions.

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LRFD From the ASCE/SEI 7 load combinations for strength design, the earthquake load is:

ASD From the ASCE/SEI 7 load combinations for strength design, the earthquake load is:

vr  Eh

vr  Eh  0.7QE

 QE  1.0  0.642 klf 

 0.7 1.0  0.642 klf 

 0.642 klf

 0.449 klf

The wind load is:

The wind load is:

vr  1.0W

vr  0.6W

 1.0  0.341 klf 

 0.6  0.341 klf 

 0.341 klf

 0.205 klf

From the SDI Diaphragm Design Manual (SDI, 2015), the nominal connection shear strength is: vn = 5.38 klf. Calculate the available strengths. LRFD Connection Strength (same for earthquake or wind) (SDI, 2015)

ASD Connection Strength (same for earthquake or wind) (SDI, 2015)

vn  0.50  5.38 klf 

vn 5.38 klf  3.25   1.66 klf  0.449 klf o.k.

 2.69 klf  0.642 klf o.k.

Check diaphragm flexibility. From the SDI Diaphragm Design Manual (SDI, 2015): K1  0.318 ft 1 K 2  870 kip/in. K 3  2,380 kip/in. K 4  3.54

K2   G'     K3  K 4  3K1 s  870 kip/in.      2,380 kip/in.  0.318   3.54  3  10 ft      ft   2, 450 kip/in.

Calculate seismic loading on the diaphragm.

105 kips 210 ft  0.500 klf

w

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Calculate the maximum diaphragm deflection. 

wL2 8bG 

 0.500 klf  210 ft 2  8  75 ft  2, 450 kip/in.  0.0150 in.

Story drift = 0.210 in. (from computer output) The diaphragm deflection is less than two times the story drift; therefore, the diaphragm is considered rigid in accordance with ASCE/SEI 7, Section 12.3.1.3. By inspection, the floor diaphragm will also be rigid in the E-W direction. Horizontal Shear Distribution and Torsion The seismic forces to be applied in the frame analysis in each direction, including the effect of accidental torsion, in accordance with ASCE/SEI 7, Section 12.8.4, are shown in Tables III-8 and III-9. Table III-8 Horizontal Shear Distribution including Accidental Torsion—Grids 1 and 8 Load on Frame Load to Grids 1 and 8 Total Fy Accidental Torsion kips % kips % kips kips Roof 64.8 50 32.4 5 3.24 35.6 4th 146 50 73.0 5 7.30 80.3 3rd 97.2 50 48.6 5 4.86 53.5 2nd 47.3 50 23.7 5 2.37 26.1 Base 196

Table III-9 Horizontal Shear Distribution including Accidental Torsion—Grids A and F Load on Frame Load to Grids A and F Total Fy Accidental Torsion kips % kips % kips kips Roof 64.8 50 32.4 5 3.24 35.6 4th 146 50 73.0 5 7.30 80.3 3rd 97.2 50 48.6 5 4.86 53.5 2nd 47.3 50.81 24.0 5 2.37 26.4 Base 196 1

Note: In this example, Grids A and F have both been conservatively designed for the slightly higher load on Grid A due to the atrium opening. The increase in load is calculated Table III-10.

I II Base

Area, ft2 25,500 841 24,700

Table III-10 Mass, kips 2,170 71.5 2,100

y-dist, ft 60.5 90.5

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My , kip-ft 131,000 6,470 125,000

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125, 000 kip-ft 2,100 kips  59.5 ft

y

100% 

121 ft  59.5 ft  121 ft

 50.8%

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MOMENT FRAME MODEL Grids 1 and 8 were modeled in conventional structural analysis software as two-dimensional models. The secondorder option in the structural analysis program was not used. Rather, for illustration purposes, second-order effects are calculated separately, using the “Approximate Second-Order Analysis” method described in AISC Specification Appendix 8. The column and beam layouts for the moment frames follow. Although the frames on Grids A and F are the same, slightly heavier seismic loads accumulate on Grid F after accounting for the atrium area and accidental torsion. The models are half-building models. The frame was originally modeled with W1482 interior columns and W2144 non-composite beams, but was revised because the beams and columns did not meet the strength requirements. The W1482 column size was increased to a W1490 and the W2144 beams were upsized to W2455 beams. Minimum composite studs are specified for the beams (corresponding to Qn = 0.25FyAs). Since the span does not exceed 30 ft, the ductility requirement is met per AISC Specification Commentary Section I3.2d.1. The beams were modeled with a stiffness of Ieq = Is. The frame was checked for both wind and seismic story drift limits. Based on the results on the computer analysis, the frame meets the L/400 drift criterion for a 10-year wind (0.7W) indicated in ASCE/SEI 7, Commentary Section CC.2.2. In addition, the frame meets the 0.025hsx allowable story drift limit given in ASCE/SEI 7, Table 12.12-1, for Risk Category II. All of the vertical loads on the frame were modeled as point loads on the frame. The dead load and live load are shown in the load cases that follow. The wind, seismic and notional loads from leaning columns are modeled and distributed 1/14 to exterior columns and 1/7 to the interior columns. This approach minimizes the tendency to accumulate too much load in the lateral system nearest an externally applied load. Also shown in the following models are the remainder of the half-building model gravity loads from the interior leaning columns accumulated in a single leaning column which was connected to the frame portion of the model with pinned ended links. Because the second-order analyses that follow will use the “Approximate Second-Order Analysis” (amplified first-order) approach given in the AISC Specification Appendix 8, the inclusion of the leaning column is unnecessary, but serves to summarize the leaning column loads and illustrate how these might be handled in a full second-order analysis. See “A Practical Approach to the ‘Leaning’ Column” (Geschwindner, 1994). There are five lateral load cases. Two are the wind load and seismic load, per the previous discussion. In addition, notional loads of Ni = 0.002Yi were established. The model layout, nominal dead, live, and snow loads with associated notional loads, wind loads and seismic loads are shown in Figures III-15 through III-23. The same modeling procedures were used in the braced frame analysis. construction, they should not be fixed in the analysis.

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If column bases are not fixed in

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Fig. III-15. Frame layout—Grid A and F.

Fig. III-16. Nominal dead loads—Grid A and F.

Fig. III-17. Notional dead loads—Grid A and F.

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Fig. III-18. Nominal live loads—Grid A and F.

Fig. III-19. Notional live loads—Grid A and F.

Fig. III-20. Nominal snow loads—Grid A and F.

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Fig. III-21. Notional snow loads—Grid A and F.

Fig. III-22. Nominal wind loads (1.0W)—Grid A and F.

Fig. III-23. Seismic loads (1.0QE)—Grid A and F.

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CALCULATION OF REQUIRED STRENGTH—THREE METHODS Three methods for checking one of the typical interior column designs at the base of the building are presented below. All three of the methods presented require a second-order analysis (either direct via computer analysis techniques or by amplifying a first-order analysis). A fourth method called the “First-Order Analysis Method” is also an option. This method does not require a second-order analysis; however, this method is not presented below. For additional guidance on applying any of these methods, see the discussion in AISC Manual Part 2 titled Required Strength, Stability, Effective Length, and Second-Order Effects. GENERAL INFORMATION FOR ALL THREE METHODS Seismic load combinations controlled over wind load combinations in the direction of the moment frames in the example building. The frame analysis was run for all LRFD and ASD load combinations; however, only the controlling combinations have been illustrated in the following examples. A lateral load of 0.2% of gravity load was included for all gravity-only load combinations per AISC Manual Part 2. The second-order analysis for all of the following examples were carried out by doing a first-order analysis and then amplifying the results to achieve a set of second-order design forces using the approximate second-order analysis procedure from AISC Specification Appendix 8.

METHOD 1—DIRECT ANALYSIS METHOD Design for stability by the direct analysis method is found in AISC Specification Chapter C. This method requires that both the flexural and axial stiffness are reduced and that 0.2% notional lateral loads are applied in the analysis to account for geometric imperfections and inelasticity, per AISC Specification Section C2.2b(a). Any general secondorder analysis method that considers both P- and P- effects is permitted. The amplified first-order analysis method of AISC Specification Appendix 7 is also permitted provided that the B1 and B2 factors are based on the reduced flexural and axial stiffnesses. A summary of the axial loads, moments and first floor drifts from the firstorder analysis is shown in the following. The floor diaphragm deflection in the east-west direction was previously determined to be very small and will thus be neglected in these calculations. Second-order member forces are determined using the approximate procedure of AISC Specification Appendix 8. It was assumed, subject to verification, that B2 is less than 1.7 for each load combination; therefore, per AISC Specification Section C2.2b(d), the notional loads were applied to the gravity-only load combinations. The required seismic load combinations, as given in ASCE/SEI 7, Section 12.4, were derived previously.

LRFD 1.23D  1.0QE  0.5 L  0.2 S (Controls columns and beams)

ASD 1.01D  0.525QE  0.75 L  0.75S (Controls columns and beams)

From a first-order analysis with notional loads where appropriate and reduced stiffnesses:

From a first-order analysis with notional loads where appropriate and reduced stiffnesses:

For interior column design

For interior column design

Pu  317 kips M u1  148 kip-ft (from first-order analysis) M u 2  233 kip-ft (from first-order analysis)

Pa  295 kips M a1  77.9 kip-ft M a 2  122 kip-ft

First story drift with reduced stiffnesses = 0.718 in.

First story drift with reduced stiffnesses = 0.377 in.

Note: For ASD, ordinarily the second-order analysis must be carried out under 1.6 times the ASD load combinations and the results must be divided by 1.6 to obtain the required strengths. For this example, second-order analysis by the approximate B1-B2 analysis method is used. This method incorporates the 1.6 multiplier directly in the B1 and B2 amplifiers, such that no other modification is needed.

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The required second-order flexural strength, Mr, and required axial strength, Pr, are determined as follows. For typical interior columns, the gravity-load moments are approximately balanced, therefore, Mnt = 0 kip-ft. Calculate the amplified forces and moments in accordance with AISC Specification Appendix 8 at the ground floor. The required second-order flexural strength is determined as follows: M r  B1M nt  B2 M lt

(Spec. Eq. A-8-1)

Determine B1

Per AISC Specification Appendix 8, Section 8.2.1, note that for members subject to axial compression, B1 may be calculated based on the first-order estimate; therefore: Pr  Pnt  Plt

where Pr = required second-order axial strength using LRFD or ASD load combinations From AISC Specification Appendix 8, Section 8.2.1, the B1 multiplier for the W1490 column is determined as follows: LRFD

ASD

Cm 1 B1  P 1 r Pe1

(Spec. Eq. A-8-3)

Cm 1 B1  P 1 r Pe1

(Spec. Eq. A-8-3)

where Pr  317 kips (from first-order computer analysis)

where Pr  295 kips (from first-order computer analysis)

I x  999 in.4 b  1.0 (to be verified per Spec. Section C2.3(b))

I x  999 in.4 b  1.0 (to be verified per Spec. Section C2.3(b))

  1.0

  1.6

As discussed in AISC Specification Appendix 8, Section 8.2.1, EI *  0.8b EI when using the direct analysis method.

Pe1 



2 EI *

(Spec. Eq. A-8-5)

 Lc1 2 2  0.8 1.0  29, 000 ksi   999 in.4 

1.0 13.5 ft 12 in./ft    8, 720 kips

Cm  0.6  0.4  M1 M 2 

2

As discussed in AISC Specification Appendix 8, Section 8.2.1, EI *  0.8b EI when using the direct analysis method.

Pe1 



2 EI *

1.0 13.5 ft 12 in./ft    8, 720 kips (Spec. Eq. A-8-4)

(Spec. Eq. A-8-5)

 Lc1 2 2  0.8 1.0  29, 000 ksi   999 in.4 

Cm  0.6  0.4  M1 M 2 

2

(Spec. Eq. A-8-4)

 0.6  0.4 148 kip-ft 233 kip-ft 

 0.6  0.4  77.9 kip-ft 122 kip-ft 

 0.346

 0.345

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LRFD 0.346 B1  1 1.0  317 kips  1 8, 720 kips  0.359  1

ASD 0.345 B1  1 1.6  295 kips  1 8, 720 kips  0.365  1

Therefore, use B1 = 1

Therefore, use B1 = 1

Determine B2 LRFD  2, 250 kips (gravity load in moment frame)

Pmf

ASD  2, 090 kips (gravity load in moment frame)

Pmf

Pstory  5, 440 kips (from computer output)

Pstory  5,120 kips (from computer output)

H 

H 

= 0.718 in. (from computer output)  1.0

 Pmf RM  1  0.15   Pstory

  

(Spec. Eq. A-8-8)

= 0.377 in. (from computer output)  1.6

 Pmf RM  1  0.15   Pstory

 2, 250 kips   1  0.15    5, 440 kips   0.938

  

(Spec. Eq. A-8-8)

 2, 090 kips   1  0.15    5,120 kips   0.939

From previous seismic force distribution calculations:

From previous seismic force distribution calculations:

H  1.0QE (Lateral)

H  0.525QE

(Lateral)

 1.0 196 kips 

 0.525 196 kips 

 196 kips

 103 kips

Pe story  RM

HL H

  0.938 

(Spec. Eq. A-8-7)

Pe story  RM

196 kips 13.5 ft 12 in./ft 

  0.939 

0.718 in.

1 1 Pstory 1 Pe story

(Spec. Eq. A-8-7)

103 kips 13.5 ft 12 in./ft  0.377 in.

 41, 600 kips

 41,500 kips

B2 

HL H

(Spec. Eq. A-8-6)

B2 

1 1 Pstory 1 Pe story

(Spec. Eq. A-8-6)

1 1 1.6  5,120 kips  1 41, 600 kips  1.25  1

1 1 1.0  5, 440 kips  1 41,500 kips  1.15  1





Because B2 < 1.7, it is verified that it was unnecessary to add the notional loads to the lateral loads for this load combination.

Because B2 < 1.7, it is verified that it was unnecessary to add the notional loads to the lateral loads for this load combination.

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Calculate amplified moment and axial load From AISC Specification Equation A-8-1, the required second-order flexural strength is determined as follows: LRFD

ASD

M r  B1M nt  B2 M lt

M r  B1M nt  B2 M lt

 1.0  0 kip-ft   1.15  233 kip-ft 

 1.0  0 kip-ft   1.25 122 kip-ft 

 268 kip-ft

 153 kip-ft

The required second-order axial strength is determined using AISC Specification Equation A-8-2 as follows. Note, for a long frame, such as this one, the change in load to the interior columns associated with lateral load is negligible. LRFD Pnt  317 kips (from computer analysis)

ASD Pnt  295 kips (from computer analysis)

Pr  Pnt  B2 Plt

Pr  Pnt  B2 Plt

 317 kips  1.15  0 kips 

 295 kips  1.25  0 kips 

 317 kips

 295 kips

Note the flexural and axial stiffness of all members in the moment frame were reduced using 0.8E in the computer analysis. Check that the flexural stiffness was adequately reduced for the analysis per AISC Specification Section C2.3(b)(1). LRFD

  1.0 Pr  317 kips

ASD

  1.6 Pr  295 kips

Because the W1490 column is nonslender:

Because the W1490 column is nonslender:

Pns  Fy Ag

Pns  Fy Ag



  50 ksi  26.5 in.2





  50 ksi  26.5 in.2

 1,330 kips

 1,330 kips

Pr 1.0  317 kips   1,330 kips Pns  0.238

Pr 1.6  295 kips   1,330 kips Pns  0.355

Because Pr/Pns  0.5:

Because Pr/Pns  0.5:

b  1.0



b  1.0

Therefore, the previous assumption is verified.

Therefore, the previous assumption is verified.

Note: By inspection b  1.0 for all of the beams in the moment frame.

Interaction of Flexure and Axial

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From AISC Specification Section H1, interaction of flexure and axial are checked as follows. From AISC Specification Section C3, K = 1.0 using the direct analysis method, therefore: Lc  KL  1.0 13.5 ft   13.5 ft

LRFD From AISC Manual Table 6-2, for a W1490, with Lc = 13.5 ft:

ASD From AISC Manual Table 6-2, for a W1490, with Lc = 13.5 ft:

Pc  c Pn  1, 040 kips

Pc 

From AISC Manual Table 6-2, for a W1490, with Lb = 13.5 ft:

From AISC Manual Table 6-2, for a W1490, with Lb = 13.5 ft:

M cx  b M nx  574 kip-ft

M cx 

Pr 317 kips  Pc 1, 040 kips  0.305

Pr 295 kips  Pc 690 kips  0.428

Because

Pn b  690 kips

M nx b  382 kip-ft

Pr  0.2 , use AISC Specification Equation Pc

Because

Pr  0.2 , use AISC Specification Equation Pc

H1-1a:

H1-1a:

Pr  8   M rx M ry        1.0 Pc  9   M cx M cy 

Pr  8   M rx M ry     Pc  9   M cx M cy

  8   268 kip-ft 0.305      0   1.0 9 574 kip-ft    0.720  1.0 o.k.

  8   153 kip-ft 0.428      0   1.0  9   382 kip-ft  0.784  1.0 o.k.

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METHOD 2—EFFECTIVE LENGTH METHOD Required strengths of frame members must be determined from a second-order analysis. In this example, the second-order analysis is performed by amplifying the axial forces and moments in members and connections from an approximate analysis using the provisions of AISC Specification Appendix 8. The available strengths of compression members are calculated using effective length factors computed from a sidesway stability analysis. A first-order frame analysis is conducted using the load combinations for LRFD or ASD. A minimum lateral load (notional load) equal to 0.2% of the gravity loads is included for any gravity-only load combination as summarized in AISC Manual Part 2 titled “Required Strength, Stability, Effective Length, and Second-Order Effects.” The required load combinations are given in ASCE/SEI 7. A summary of the axial loads, moments and 1st floor drifts from the first-order computer analysis is shown below. The floor diaphragm deflection in the east-west direction was previously determined to be very small and will thus be neglected in these calculations. LRFD 1.23D  1.0QE  0.5 L  0.2 S (Controls columns and beams)

ASD 1.01D  0.525QE  0.75 L  0.75S (Controls columns and beams)

For interior column design:

For interior column design:

Pu  317 kips M u1  148 kip-ft (from first-order analysis) M u 2  233 kip-ft (from first-order analysis)

Pa  295 kips M a1  77.9 kip-ft (from first-order analysis) M a 2  122 kip-ft (from first-order analysis)

First-order story drift = 0.575 in.

First-order story drift = 0.302 in.

The required second-order flexural strength, Mr, and axial strength, Pr, are calculated as follows. For typical interior columns, the gravity load moments are approximately balanced; therefore, Mnt = 0 kip-ft. Calculate the amplified forces and moments in accordance with AISC Specification Appendix 8 at the ground floor. The required second-order flexural strength is determined as follows: M r  B1M nt  B2 M lt

(Spec. Eq. A-8-1)

Determine B1

Per AISC Specification Appendix 8, Section 8.2.1, note that for members subject to axial compression, B1 may be calculated based on the first-order estimate; therefore: Pr  Pnt  Plt

where Pr = required second-order axial strength using LRFD or ASD load combinations From AISC Specification Appendix 8, Section 8.2.1, the B1 multiplier for the W1490 column is determined as follows: LRFD

Cm 1 B1  P 1 r Pe1

ASD (Spec. Eq. A-8-3)

Cm 1 B1  P 1 r Pe1

LRFD

(Spec. Eq. A-8-3)

ASD

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where Pr  317 kips (from first-order computer analysis)

where Pr  295 kips (from first-order computer analysis)

I x  999 in.4

I x  999 in.4

b  1.0 (to be verified per Spec. Section C2.3(b))

b  1.0 (to be verified per Spec. Section C2.3(b))

  1.0

  1.6

Pe1 



2 EI *

(Spec. Eq. A-8-5)

 Lc1 2 2  29, 000 ksi   999 in.4 

1.0 13.5 ft 12 in./ft    10,900 kips

Cm  0.6  0.4  M1 M 2 

Pe1 



2

2 EI *

 Lc1 2 2  29, 000 ksi   999 in.4 

1.0 13.5 ft 12 in./ft    10,900 kips (Spec. Eq. A-8-4)

Cm  0.6  0.4  M1 M 2 

(Spec. Eq. A-8-5)

2

(Spec. Eq. A-8-4)

 0.6  0.4 148 kip-ft 233 kip-ft 

 0.6  0.4  77.9 kip-ft 122 kip-ft 

 0.346

 0.345

0.346 1 1.0  317 kips  1 10,900 kips  0.356  1

0.345 1 1.6  295 kips  1 10, 900 kips  0.361  1

B1 

B1 

Therefore, use B1 = 1

Therefore, use B1 = 1

Determine B2 Pmf

LRFD  2, 250 kips (gravity load in moment frame)

Pmf

ASD  2, 090 kips (gravity load in moment frame)

Pstory  5, 440 kips (from computer output)

Pstory  5,120 kips (from computer output)

H 

H 

= 0.575 in. (from computer output)  1.0

 Pmf  RM  1  0.15    Pstory   2, 250 kips   1  0.15    5, 440 kips 

(Spec. Eq. A-8-8)

 0.938

= 0.302 in. (from computer output)  1.6

 Pmf  RM  1  0.15    Pstory   2, 090 kips   1  0.15    5,120 kips 

(Spec. Eq. A-8-8)

 0.939

From previous seismic force distribution calculations:

From previous seismic force distribution calculations:

H  1.0QE (Lateral)

H  0.525QE (Lateral)

 1.0 196 kips 

 0.525 196 kips 

 196 kips

 103 kips

LRFD

ASD

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HL (Spec. Eq. A-8-7) H 196 kips 13.5 ft 12 in./ft   0.938 0.575 in.  51,800 kips

Pe story  RM

B2 

1 1 Pstory 1 Pe story

(Spec. Eq. A-8-6)

HL (Spec. Eq. A-8-7) H 103 kips 13.5 ft 12 in./ft   0.939 0.302 in.  51,900 kips

Pe story  RM

B2 

1 1 Pstory 1 Pe story

(Spec. Eq. A-8-6)

1 1 1.6  5,120 kips  1 51,900 kips  1.19  1

1 1 1.0  5, 440 kips  1 51,800 kips  1.12  1





Note, B2 < 1.5, therefore use of the effective length method is acceptable per AISC Specification Appendix 7, Section 7.2.1(b).

Note, B2 < 1.5, therefore use of the effective length method is acceptable per AISC Specification Appendix 7, Section 7.2.1(b).

Calculate amplified moment and axial load From AISC Specification Equation A-8-1, the required second-order flexural strength is determined as follows: LRFD

ASD

M r  B1M nt  B2 M lt

M r  B1M nt  B2 M lt

 1 0 kip-ft   1.12  233 kip-ft 

 1 0 kip-ft   1.19 122 kip-ft 

 261 kip-ft

 145 kip-ft

The required second-order axial strength is determined using AISC Specification Equation A-8-2 as follows. Note, for a long frame, such as this one, the change in load to the interior columns associated with lateral load is negligible. LRFD Pnt  317 kips (from computer analysis)

ASD Pnt  295 kips (from computer analysis)

Pr  Pnt  B2 Plt

Pr  Pnt  B2 Plt

 317 kips  1.12  0 kips 

 295 kips  1.19  0 kips 

 317 kips

 295 kips

Determine the Controlling Effective Length For out-of-plane buckling in the moment frame, Ky = 1.0; therefore: K y Ly  1.0 13.5 ft   13.5 ft

For in-plane buckling in the moment frame, use the story stiffness procedure from AISC Specification Commentary Appendix 7 to determine Kx.

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 Pstory K2    RM Pr

 2 EI    H    2 EI    H      2    2      L   HL   L   1.7 H col L 

(Spec. Eq. C-A-7-5)

Simplifying and substituting terms previously calculated results in:  Pstory  Pe  ratio   ratio  Kx       Pe   R P H   1.7 H   M  r  where Pe  Pe1

ratio 

H L ASD

LRFD Pe  Pe1

Pe  Pe1

 10, 900 kips

ratio  

 10, 900 kips

H L

ratio 

0.575 in. 13.5 ft 12 in./ft 



 0.00355

H L 0.302 in. 13.5 ft 12 in./ft 

 0.00186

 5, 440 kips   10,900 kips  0.00355  Kx       0.938   317 kips   196 kips 

 5,120 kips   10,900 kips  0.00186  Kx       0.939   295 kips   103 kips  

0.00186   1.7 103 kips  



0.00355   1.7 196 kips  

10,900 kips  

10,900 kips  

 1.91  0.340

 1.90  0.341

Therefore, use Kx = 1.90.

Therefore, use Kx = 1.91.

From AISC Manual Table 4-1a, for a W1490:

From AISC Manual Table 4-1a, for a W1490:

rx ry  1.66

rx ry  1.66

Lcy eq  

KLx rx ry

(from Manual Eq. 4-1)

1.90 13.5 ft 

1.66  15.5 ft

Because Lcy eq  Lcy , use Lc = 15.5 ft.

Lcy eq  

KLx rx ry

(from Manual Eq. 4-1)

1.9113.5 ft 

1.66  15.5 ft

Because Lcy eq  Lcy , use Lc = 15.5 ft.

Interaction of Flexure and Axial From AISC Specification Section H1, interaction of flexure and axial are checked as follows:

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LRFD From AISC Manual Table 6-2, for a W1490, with Lc = 15.5 ft: Pc  c Pn

Pn c  660 kips

Pc 

 990 kips

From AISC Manual Table 6-2, for a W1490, with Lb = 13.5 ft: M cx  b M nx

M nx b  382 kip-ft

Pr 317 kips  Pc 990 kips  0.320

Pr 295 kips  Pc 660 kips  0.447

Pr  0.2 , use AISC Specification Equation Pc

Because

Pr  0.2 , use AISC Specification Equation Pc

H1-1a:

H1-1a: Pr  8   M rx M ry     Pc  9   M cx M cy

From AISC Manual Table 6-2, for a W1490, with Lb = 13.5 ft: M cx 

 574 kip-ft

Because

ASD From AISC Manual Table 6-2, for a W1490, with Lc = 15.5 ft:

   1.0 

 8   261 kip-ft  0.320       1.0  9   574 kip-ft  0.724  1.0 o.k.

Pr  8   M rx M ry     Pc  9   M cx M cy

   1.0 

 8   145 kip-ft  0.447       1.0  9   382 kip-ft  0.784  1.0 o.k.

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METHOD 3—SIMPLIFIED EFFECTIVE LENGTH METHOD A simplification of the effective length method using a method of second-order analysis based upon drift limits and other assumptions is described in Part 2 of the AISC Manual titled “Simplified Determination of Required Strength.” A first-order frame analysis is conducted using the load combinations for LRFD or ASD. A minimum lateral load (notional load) equal to 0.2% of the gravity loads is included for all gravity-only load combinations. The floor diaphragm deflection in the east-west direction was previously determined to be very small and will thus be neglected in these calculations. LRFD 1.23D  1.0QE  0.5 L  0.2 S (Controls columns and beams)

ASD 1.01D  0.525QE  0.75 L  0.75S (Controls columns and beams)

For interior column design:

For interior column design:

Pu  317 kips M u1  148 kip-ft (from first-order analysis) M u 2  233 kip-ft (from first-order analysis)

Pa  295 kips M a1  77.9 kip-ft (from first-order analysis) M a 2  122 kip-ft (from first-order analysis)

First-order first story drift = 0.575 in.

First-order first story drift = 0.302 in.

Calculate the amplified forces and moments in accordance with AISC Manual Part 2 at the ground floor. The following steps are executed.

LRFD

ASD

Step 1:

Step 1:

Lateral load = 196 kips

Lateral load = 103 kips

Deflection due to first-order elastic analysis

Deflection due to first-order elastic analysis

 = 0.575 in., between first and second floor

 = 0.302 in., between first and second floor

Floor height = 13.5 ft

Floor height = 13.5 ft

Drift ratio 

13.5 ft 12 in./ft 

Drift ratio 

0.575 in.

 282

13.5 ft 12 in./ft  0.302 in.

 536

Step 2:

Step 2:

Design story drift limit = H/400

Design story drift limit = H/400

 282  Adjusted lateral load    196 kips   400   138 kips

 536  Adjusted lateral load    103 kips   400   138 kips

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Step 3:

LRFD

ASD Step 3: (for an ASD design the ratio must be multiplied by 1.6)

 total story load  Load ratio = 1.0     lateral load   5, 440 kips  = 1.0     138 kips 

 total story load  Load ratio = 1.6     lateral load   5,120 kips  = 1.6     138 kips 

= 39.4

= 59.4

From AISC Manual Table 2-1:

From AISC Manual Table 2-1:

B2 = 1.1

B2 = 1.2

Which matches the value obtained in Method 2 to the two significant figures of the table

Which matches the value obtained in Method 2 to the two significant figures of the table

Note: Intermediate values are not interpolated from the table because the precision of the table is two significant digits. Additionally, the design story drift limit used in Step 2 need not be the same as other strength or serviceability drift limits used during the analysis and design of the structure. Step 4: Multiply all the forces and moment from the first-order analysis by the value of B2 obtained from the table. This presumes that B1 is less than or equal to B2, which is usually the case for members without transverse loading between their ends. LRFD

ASD

Step 5:

Step 5:

Since the selection is in the shaded area of the chart, (B2  1.1), use K = 1.0.

Since the selection is in the unshaded area of the chart (B2 > 1.1), the effective length factor, K, must be determined through analysis. From previous analysis, use an effective length of 15.5 ft.

Multiply both sway and nonsway moments by B2.

Multiply both sway and nonsway moments by B2.

M r  B2  M nt  M lt 

M r  B2  M nt  M lt 

 1.1 0 kip-ft  233 kip-ft 

 1.2  0 kip-ft  122 kip-ft 

 256 kip-ft

 146 kip-ft

Pr  B2  Pnt  Plt 

Pr  B2  Pnt  Plt   1.1 317 kips  0 kips 

 1.2  295 kips  0 kips 

 349 kips

 354 kips

From AISC Manual Table 6-2, for a W1490, with Lc = 13.5 ft:

From AISC Manual Table 6-2, for a W1490, with Lc = 15.5 ft:

Pc  c Pn

Pc 

 1, 040 kips

Pn c  660 kips

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LRFD From AISC Manual Table 6-2, for a W1490, with Lb = 13.5 ft: M cx  b M nx

M nx b  382 kip-ft

M cx 

 574 kip-ft

Pr 349 kips  Pc 1,040 kips  0.336

Because

Pr 354 kips  Pc 660 kips  0.536

Pr  0.2, use AISC Specification Equation Pc

H1-1a: Pr  8   M rx M ry     Pc  9   M cx M cy

ASD From AISC Manual Table 6-2, for a W1490, with Lb = 13.5 ft:

Because

Pr  0.2, use AISC Specification Equation Pc

H1-1a:    1.0 

  8   256 kip-ft 0.336      0   1.0 9 574 kip-ft    0.732  1.0 o.k.

Pr  8   M rx M ry     Pc  9   M cx M cy

   1.0 

  8   146 kip-ft 0.536      0   1.0  9   382 kip-ft  0.876  1.0 o.k.

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BEAM ANALYSIS IN THE MOMENT FRAME The controlling load combinations for the beams in the moment frames are shown in Tables III-11 and III-12, and evaluated for the second floor beam. The dead load, live load and seismic moments were taken from a computer analysis. These tables summarizes the calculation of B2 for the stories above and below the second floor. Table III-11 Summary of B2 Calculation for Controlling Load Combination—First to Second Floor 1st – 2nd LRFD Combination ASD Combination 1 ASD Combination 2 1.23D + 1.0QE + 0.5L + 0.2S 1.02D + 0.7QE 1.01D + 0.525QE + 0.75L + 0.75S 196 kips 137 kips 103 kips H 13.5 ft 13.5 ft 13.5 ft L 0.575 in. 0.402 in. 0.302 in. H 2,250 kips 1,640 kips 2,090 kips Pmf 0.938 0.937 0.939 RM 51,800 kips 51,700 kips 51,900 kips Pe story 5,440 kips 3,920 kips 5,120 kips Pstory B2 1.12 1.14 1.19

Table III-12 Summary of B2 Calculation for Controlling Load Combination—Second to Third Floor 2nd – 3rd LRFD Combination ASD Combination 1 ASD Combination 2 1.23D + 1.0QE + 0.5L + 0.2S 1.02D + 0.7QE 1.01D + 0.525QE + 0.75L + 0.75S 170 kips 119 kips 89.3 kips H 13.5 ft 13.5 ft 13.5 ft L 0.728 in. 0.509 in. 0.382 in. H 1,590 kips 1,160 kips 1,490 kips Pmf 0.938 0.937 0.939 RM 35,500 kips 35,500 kips 35,600 kips Pe story 3,840 kips 2,770 kips 3,660 kips Pstory B2 1.12 1.14 1.20

For beam members, the larger of the B2 values from the story above or below is used.

From computer output at the controlling beam: M dead

 153 kip-ft

M live  80.6 kip-ft M snow  0 kip-ft M earthquake  154 kip-ft

LRFD B2 M lt  1.12 154 kip-ft 

ASD Combination 1:

 172 kip-ft

B2 M lt  1.14 154 kip-ft 

1.23 153 kip-ft   1.0 172 kip-ft   Mu =     0.5  80.6 kip-ft    400 kip-ft

 176 kip-ft M a =1.02 153 kip-ft   0.7 176 kip-ft   279 kip-ft

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LRFD

ASD Combination 2: B2 M lt  1.20 154 kip-ft   185 kip-ft

1.01153 kip-ft   0.525 185kip-ft   Ma =     0.75  80.6 kip-ft    312 kip-ft Calculate Cb for W2455 beam with compression in the bottom flange braced at 10 ft on center. LRFD For load combination 1.23D + 1.0QE + 0.5L + 0.2S:

ASD For load combination 1.02D + 0.7QE:

From AISC Manual Table 6-2 with Lb = 0 ft (fully braced):

From AISC Manual Table 6-2 with Lb = 0 ft (fully braced):

b M n  503 kip-ft

Mn  334 kip-ft b

Cb = 1.86 (from computer output)

Cb = 1.86 (from computer output)

From AISC Manual Table 6-2 with Lb = 10 ft:

From AISC Manual Table 6-2 with Lb = 10 ft:

b M n Cb  b M p

Mp Mn Cb  b b

 386 kip-ft 1.86   718 kip-ft  503 kip-ft Therefore: M n  503 kip-ft  400 kip-ft

 257 kip-ft 1.86   478 kip-ft  334 kip-ft Therefore:

o.k.

Mn  334 kip-ft  279 kip-ft 

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LRFD

ASD For load combination 1.01D + 0.525QE + 0.75L: From AISC Manual Table 6-2 with Lb = 0 ft (fully braced): Mn  334 kip-ft b

Cb = 2.01 (from computer output) From AISC Manual Table 6-2 with Lb = 10 ft : Mp Mn Cb  b b

 257 kip-ft  2.01  517 kip-ft  334 kip-ft Therefore: Mn  334 kip-ft  312 kip-ft  From AISC Manual Table 6-2, a W2455 has a design shear strength of 252 kips and an Ix of 1,350 in.4

o.k.

From AISC Manual Table 6-2, a W2455 has an allowable shear strength of 167 kips and an Ix of 1,350 in.4

The moments and shears on the roof beams due to the lateral loads were also checked but do not control the design. The connections of these beams can be designed by one of the techniques illustrated in the Chapter IIB of the design examples.

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BRACED FRAME ANALYSIS The braced frames at Grids 1 and 8 were analyzed for the required load combinations. The stability design requirements from Chapter C were applied to this system. The model layout is shown in Figure III-24. The nominal dead, live, and snow loads with associated notional loads, wind loads and seismic loads are shown in Figures III-25 and III-26.

Fig. III-24. Braced frame layout—Grid 1 and 8.

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(a) Nominal dead loads

(b) Notional dead loads

(c) Nominal live loads

(d) Notional live loads Fig. III-25. Dead and live loads.

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(a) Nominal snow loads

(b) Notional snow loads

(c) Wind loads (1.0W)

(d) Seismic loads (1.0QE)

Fig. III-26. Snow, wind and seismic loads.

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Second-order analysis by amplified first-order analysis In the following, the approximate second-order analysis method from AISC Specification Appendix 8 is used to account for second-order effects in the braced frames by amplifying the axial forces in members and connections from a first-order analysis. A first-order frame analysis is conducted using the load combinations for LRFD and ASD. From this analysis the critical axial loads, moments and deflections are obtained. A summary of the axial loads and first floor drifts from the first-order computer analysis is shown below. The floor diaphragm deflection in the north-south direction was previously determined to be very small and will thus be neglected in these calculations. The required seismic load combinations, as given in ASCE/SEI 7, Section 12.4, were derived previously.

LRFD 1.23D  1.0QE  0.5 L  0.2 S (Controls columns and beams)

ASD 1.01D  0.525QE  0.75 L  0.75S (Controls columns and beams)

From first-order analysis.

From first-order analysis.

For interior column design:

For interior column design:

Pnt  236 kips Plt  146 kips

Pnt  219 kips Plt  76.6 kips

The moments are negligible.

The moments are negligible.

First-order first story drift = 0.211 in.

First-order first story drift = 0.111 in.

The required second-order axial strength, Pr, is computed as follows:

LRFD Pr  Pnt  B2 Plt

ASD (Spec. Eq. A-8-2)

Determine B2. B2 

1 1 Pstory 1 Pe story

HL H

(Spec. Eq. A-8-2)

Determine B2. (Spec. Eq. A-8-6)

Pstory  5, 440 kips (previously calculated)

Pe story  RM

Pr  Pnt  B2 Plt

(Spec. Eq. A-8-7)

where H = 196 kips (from previous calculations) H = 0.211 in. (from computer output) RM = 1.0 for braced frames

B2 

1 1 Pstory 1 Pe story

(Spec. Eq. A-8-6)

Pstory  5,120 kips (previously calculated)

Pe story  RM

HL H

(Spec. Eq. A-8-7)

where H = 103 kips (from previous calculations) H = 0.111 in. (from computer output) RM = 1.0 for braced frames

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LRFD

Pe story

 196 kips 13.5 ft 12 in./ft    1.0    0.211 in.    150, 000 kips

1 1 1.0  5, 440 kips  1 150, 000 kips  1.04  1

Pe story

ASD  103 kips 13.5 ft 12 in./ft    1.0    0.111 in.    150, 000 kips

1 1 1.6  5,120 kips  1 150, 000 kips  1.06  1

B2 

B2 

Therefore, use B2 = 1.04.

Therefore, use B2 = 1.06.

Pr  Pnt  B2 Plt

(Spec. Eq. A-8-2)

Pr  Pnt  B2 Plt

 236 kips  1.04 146 kips 

 219 kips  1.06  76.6 kips 

 388 kips

 300 kips

(Spec. Eq. A-8-2)

From AISC Manual Table 6-2 for a W1253 with Lc = 13.5 ft:

From AISC Manual Table 6-2 for a W1253 with Lc = 13.5 ft:

Pc  c Pn  514 kips

Pc 

From AISC Specification Equation H1-1a:

From AISC Specification Equation H1-1a:

Pr 388 kips   1.0 Pc 514 kips  0.755  1.0 o.k.

Pr 300 kips   1.0 Pc 342 kips  0.877  1.0 o.k.

Pn c  342 kips

Note: Notice that the lower sidesway displacements of the braced frame produce much lower values of B2 than those of the moment frame. Similar results could be expected for the other two methods of analysis. Although not presented here, second-order effects should be accounted for in the design of the beams and diagonal braces in the braced frames at Grids 1 and 8.

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ANALYSIS OF DRAG STRUTS The fourth floor delivers the highest diaphragm force to the braced frames at the ends of the building: QE = 80.3 kips (from previous calculations). This force is transferred to the braced frame through axial loading of the W1835 beams at the end of the building. The gravity dead loads for the edge beams are the floor loading of 75 psf (5.50 ft) plus the exterior wall loading of 0.503 kip/ft, giving a total dead load of 0.916 kip/ft. The gravity live load for these beams is the floor loading of 80 psf (5.50 ft) = 0.440 kip/ft. The resulting midspan moments are MD = 58.0 kip-ft and ML = 27.8 kip-ft. The required seismic load combinations, as given in ASCE/SEI 7, Section 12.4, were derived previously. The controlling load combination for LRFD is 1.23D + 1.0QE + 0.5L. The controlling load combinations for ASD are 1.01D + 0.525QE + 0.75L or 1.02D + 0.7QE.

LRFD

ASD

M u  1.23M D  0.5M L

M a  1.01M D  0.75M L

 1.23  58.0 kip-ft   0.5  27.8 kip-ft 

 1.01 58.0 kip-ft   0.75  27.8 kip-ft 

 85.2 kip-ft

 79.4 kip-ft

or M a  1.02 M D  1.02  58.0 kip-ft 

Load from the diaphragm shear due to earthquake loading

 59.2 kip-ft  Load from the diaphragm shear due to earthquake loading

Fp  1.0QE

Fp  0.525QE

 1.0  80.3 kips 

 0.525  80.3 kips 

 80.3 kips

 42.2 kips

or Fp  0.7QE  0.7  80.3 kips   56.2 kips

Only the two 45-ft-long segments on either side of the brace can transfer load into the brace, because the stair opening is in front of the brace. Use AISC Specification Section H2 to check the combined bending and axial stresses.

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LRFD

ASD

80.3 kips V  2  45 ft 

42.2 kips V  2  45 ft 

 0.892 kip/ft

 0.469 kip/ft

or V 

56.2 kips 2  45 ft 

 0.624 kip/ft

From AISC Manual Table 1-1, for a W1835:

S x  57.6 in.3 LRFD The top flange bending stress is: f rbw  

ASD The top flange bending stress is:

Mu Sx

f rbw 

 85.2 kip-ft 12 in./ft 



Ma Sx

 79.4 kip-ft 12 in./ft 

57.6 in.3  16.5 ksi

3

57.6 in.  17.8 ksi

or f rbw  

Ma Sx

 59.2 kip-ft 12 in./ft 

57.6 in.3  12.3 ksi

Note: It is often possible to resist the drag strut force using the slab directly. For illustration purposes, this solution will instead use the beam to resist the force independently of the slab. The full cross section can be used to resist the force if the member is designed as a column braced at one flange only (plus any other intermediate bracing present, such as from filler beams). Alternatively, a reduced cross section consisting of the top flange plus a portion of the web can be used. Arbitrarily use the top flange and 8 times an area of the web equal to its thickness times a depth equal to its thickness, as an area to carry the drag strut component. Area  b f t f  8  t w 

2

  6.00 in. 0.425 in.  8  0.300 in.

2

 3.27 in.2

Ignoring the small segment of the beam between Grid C and D, the axial stress due to the drag strut force is:

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LRFD

fra 

80.3 kips



2 3.27 in.2

ASD

fra 



 12.3 ksi

42.2 kips



2 3.27 in.2



 6.45 ksi or

fra 

56.2 kips





2 3.27 in.2 

 8.59 ksi LRFD Using AISC Specification Section H2, assuming the top flange is continuously braced:

ASD From AISC Specification Section H2, assuming the top flange is continuously braced:

Fca  c Fy

Fca  Fy c

 0.90  50 ksi 

 50 ksi 1.67  29.9 ksi

 45.0 ksi

Fy b  50 ksi 1.67

Fcbw  b Fy

Fcbw 

 0.90  50 ksi   45.0 ksi

 29.9 ksi

f ra f rbw (from Spec. Eq. H2-1)   1.0 Fca Fcbw 12.3 ksi 17.8 ksi   0.669  1.0 o.k. 45.0 ksi 45.0 ksi

f ra f rbw   1.0 Fca Fcbw

(from Spec. Eq. H2-1)

Load Combination 1: 6.45 ksi 16.5 ksi   0.768  1.0 29.9 ksi 29.9 ksi

o.k.

Load Combination 2: 8.59 ksi 12.3 ksi   0.699  1.0 29.9 ksi 29.9 ksi

o.k.



Note: Because the drag strut load is a horizontal load, the method of transfer into the strut, and the extra horizontal load that must be accommodated by the beam end connections should be indicated on the drawings.

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PART III EXAMPLE REFERENCES

ASCE (2014), Design Loads on Structures During Construction, ASCE/SEI 37-14, American Society of Civil Engineers, Reston, VA. Geschwindner, L.F. (1994), “A Practical Approach to the Leaning Column,” Engineering Journal, AISC, Vol. 31, No. 4, pp. 141–149. SDI (2014), Floor Deck Design Manual, 1st Ed., Steel Deck Institute, Glenshaw, PA. SDI (2015), Diaphragm Design Manual, 4th Ed., Steel Deck Institute, Glenshaw, PA. SJI (2015), Load Tables and Weight Tables for Steel Joists and Joist Girders, 44th Ed., Steel Joist Institute, Forest, VA. West, M.A. and Fisher, J.M. (2003), Serviceability Design Considerations for Steel Buildings, Design Guide 3, 2nd Ed., AISC, Chicago, IL.

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